Lecture 13. Aqueous Equilibria Pt. 2.
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. 
 A quick announcement. I wanted to remind you that yesterday you would have received a PDF file of your graded exam. So take a look at the exam and if you have any questions about the grading, you should print out the entire exam and take it with you to discussion. And talk to your TA about it. If you do think there is an error, then what you need to do is print out a copy of the exam, attach a note explaining to me what the problem is, and you need to hand it to me, not the TAs. So all regrades have to be turned into me. Please do not send me an email because as you know, when it's a class of this size, it's just I can't be non-dated with PDF files and it's difficult for me to keep track of who sent what. So that's why I'm asking you to print it out, attach a note, and then turn it into me, and the deadline is two weeks from yesterday because it's two weeks from the date when you received the PDF files from me. And so within that deadline, if you turn it in, then we will take a look at it. Is that clear to everybody? So please do not send me an email. You can hand it to me after class or you can hand it to me before class. You can hand it to me during my office hours. And if I'm not there, you can even slide it out of my door as long as I get it within the two week deadline. Is that clear to everybody? So if you guys remember, we're looking at acid-base titrations and we said we're going to look at three types of acid-base titrations. Strong acid, strong base of isewersa. We looked at weak acid, strong base. Now we're going to look at the third scenario where you have a weak base versus a strong acid. So if you look at the titration, you can see that the weak base is placed in the Erlenmey flask. And that is typically in an analysis, that is the one that has the unknown concentration. We placed the strong acid in the burette. And when you start the titration now, we want to look at what the reaction would be. Remember we said that if you have an acid in a base, they can't coexist. Large amounts of acids and bases do not coexist. They will react to give you water and salt. So let's see what happens here. So if you write the complete ionic equation, and we'll start with the strong acid. A strong acid we know dissociates completely. So in reality when it's dissolved in water, you have hydronium ion and chloride ion. So that's the strong acid. Now we have the weak base. We know weak bases do not dissociate, do not react with water. They tend to be undissociated. Or in this case, they would not pick up a proton because it's a weak base. We know that when we see the word weak, it means its preference is to remain the way it is. So this is what we have at the start. Now we're going to let it go to products. Now this is the base. This is the acid. We know they're going to react to give me NH4 plus aqueous plus water plus Cl minus aqueous. And so what you end up with is this strong acid and the weak base will give you its conjugate acid and water. And if you look at this, chloride ions are there at the beginning, chloride ions are there at the end. So the chloride ions are the spectator ions. So we have only one spectator ion here, which is the chloride ion. Now if you write the net ionic equation, now we take away the spectator ions. So because remember the spectator ions don't participate in the reaction, they're just there. And so if I write the net ionic equation, what I end up with is the hydronium ion plus ammonia, which is the acid and the base, will react to give me water NH4 plus, which is the conjugate acid and water. So this is the acid. This is the base. The acid will donate a proton to the base so that you end up with NH4 plus and water. So this is the neutralization. And so if we look at this, if all of this reacted completely with this at the equivalence point or the stoichiometric point, what you'd have is water. Water is neutral plus you have NH4 plus. So now look at NH4 plus. NH4 plus is the conjugate acid of a weak base, which is NH3. So if you take NH4 plus, would you expect it to be acidic, basic, or neutral? Acidic. All right? At the equivalence point now, the pH has to be acidic. All right? And then if you write the overall reaction or the general form of the reaction, it will be where you have HCl, which is the strong acid reacting with ammonia, which is the weak base giving you NH4 plus. In fact, I'm writing the general form of the reaction. So it just turned out to be giving you salt here. All right? So if you write it in the general form, you see that what you're forming is ammonium chloride. You're forming the salt, the strong acid reacting with the base. So can everybody see that the general form or the overall reaction or sometimes they call this the molecular equation, all right, where it identifies the molecules? See that these are covalent compounds, so you can use the word molecular, all right? And this is also called the molecular equation. Okay? So now let's look at what the pH titration curve looks like. So this is what the titration curve looks like. So over here, we're measuring the pH on this side. On the y-axis, we have the pH of the analyte. Over here, we have the volume of the strong acid that's added. And you can see now the shape of this is very different. Okay? To begin with, you just have a weak base, all right? Before we add any strong acid, we have only a weak base in the flat. So we'd expect the pH to be high, all right? It's a basic solution. Now as you add a strong acid and you're neutralizing the weak base and converting it to NH4 plus, which is a weak acid, you know that the pH is going to go down. Okay? So what we see is that the pH decreases. But in this region, this is the buffer region, all right? Once again, over here in this region here, it would act as a buffer. So that's why it flattens out. The pH doesn't change dramatically there. It just kind of flattens out in plateaus because now what we have is in that mix, in that solution, you'll have NH3, which is the weak base, and you're going to have NH4 plus, which is its conjugate acid, and you have mixtures of both in substantial amounts. And so in that region, it's going to act as a buffer, all right? So that's where this flattens out. And then as you add more and more acid, now your buffer is going to get completely swamped and everything is going to get converted to NH4 plus. So at the equivalence point, now you can see all I have is NH4 plus, Cl minus, and water. Cl minus is the conjugate base of a strong acid. We know that's neutral. So Cl minus is neutral. Water is neutral, all right? And therefore, the only other species you have in there is NH4 plus, and the NH4 plus is the conjugate acid of a weak base. It's going to be acidic. And therefore, the pH is less than 7, all right? So you can see that the pH is less than 7, okay? Beyond this point, now you're swamped with a strong acid. So beyond the equivalence point, the pH is going to determine by the extra hydrochloric acid that's there, which associates completely. And you can see beyond the equivalence point, now rapidly the solution gets very acidic, all right? Now if this is the equivalence point, so you can see that in this titration, the concentrations are different, okay? In this titration, it turns out that the equivalence point is at 50 milliliters, all right? So half the equivalent point would be 25. So if you take 25 milliliters, that would correspond to that. And so at 25 milliliters, we know that we have a buffer there where the concentrations of the weak acid are equal to each other. So the concentrations are equal to each other, you know, it's a buffer. And if the concentrations are equal to each other, the ratio of acid to base would be 1. Log 1 would be what? 0. And therefore, we know pH equals pKa, all right? So here, even though we have a base and solution, this gives me the point at which pH equals pKa, all right? It's not pH equals pKa, even though we have an acid down there, some students say, well, we have a base down there, so why are we talking about kA rather than kB? But remember, pH always equals pKa, according to Henderson-Husselbalck equation, where the concentrations of acid and base are the same, all right? So here, this would be pH equals pKa. And so you don't even have to do any calculations. I ask you, what is the pH of the half-equivalence point? You just look at this table and you look for the conjugate acid-base pair that we're interested in looking at, and that would be ammonium and ammonia. And you know, pKa is 9.25, so what should this pH correspond to? 9.25, all right? Because we know that at the half-equivalence point, the pH has to equal pKa. Is that clear to everybody? So now we've looked at all three types of titrations. I did not do any calculations here, but you should be able to do the same type of calculations that we did for weak acid-strong base or strong acid-strong base. And I'd like you to try that at home. In fact, I've assigned a problem from your textbook where you get to practice that. Okay? Yes, go ahead. Can you explain what happened to the equivalent point, meaning the volume? So her question was, what happens at the equivalent point? So you have to turn to the net ionic equation. So in this titration, what's happening is the strong acid neutralizes the weak base, all right? And so essentially what happens is, when this reaction takes place, you have NH4 plus and water being formed. Okay? So when you have stoichiometric amounts of both or equivalent amounts of both, where all of this reacts completely with this and all of this reacts completely with this, so that you have no excess reagent left over, all right? Both reactants react completely. And that happens only at the equivalent point because you have to have stoichiometric amounts. You have to have the right amount of each so that you don't have any limiting reagent. You don't have any excess reagent. They react in stoichiometric amounts. So that only happens at the equivalent point, okay? So at the equivalent point, when both of the reactants have reacted completely, all that you have left behind is just the product. You don't have any excess reagent left over. So because all of them are reacted, all that you have there is just NH4 plus, water, and Cl minus, all right? The spectator ion is Cl minus. Cl minus is the conjugate base of a strong acid, and we know that that's neutral. We know water is neutral, all right? And so now we turn to NH4 plus. So those are only three species in solution. If you turn to NH4 plus, you know that that's a weak acid. It's the conjugate acid of a weak base, and therefore it's a weak acid. If it's a weak acid, it's going to be acidic, all right? And therefore the pH should be less than 7. Got it? Okay, so now we've completed looking at acid-based titrations, and now we're actually ready to move on to the last topic that we're going to look at in terms of titrations. So by now, you guys are experts at solving any equilibrium problem. And we focused, this entire quarter, we focused on equisiculabria, what happens in water, all right? And that kind of has been the focus all along. Now we're going to look at the last type of equilibria, and this deals with solubility, all right? So we're going to look at solubility product. And as all of you know, solubility deals with when a substance dissolves. So we're looking at solubility product in equisolutions, and so the word solubility is related to soluble when something dissolves, all right? So let's take something, as all of you know, we know that many, many ionic compounds dissolve easily in water, and we call them soluble salts. So if I take soluble salts. So we'll start by looking at soluble salts. And all of you know, when we use the word salts, we're talking about ionic compounds, and we said if you take ionic compounds that are soluble, what happens the instant you dissolve them in water, what happens to soluble salts? They dissolve, and when they dissolve, what does that tell you? They dissociate completely, okay? So give me an example of a really good soluble salt in ACL. We all know table salt in ACL is a very good soluble salt. So I start with an ACL, and all of you know that if I take a beaker, and I'm going to add solid sodium chloride. So you take, all of you know that ionic compounds tend to be crystalline solids, and if you take common table salt, you can see that they have nice crystalline granules. And so if you take a spoonful of that and put it in a beaker water and stir it, what's going to happen? It dissolves completely. And when we say it dissolves completely, we know that it has dissociated completely to give you Na plus and Cl minus, okay? Now what I'm going to do is I'm going to add another spoonful and stir it up. It's going to dissolve. I'm going to take another spoonful and stir it up. It'll dissolve. I'm going to keep doing this until the point comes when that solution is completely saturated with sodium chloride. So if I keep putting more and more and more sodium chloride, what I'll end up with is a really saturated solution. And the point at which this gets to be saturated, we know it's saturated because it can't take anymore. If you add another spoonful of sodium chloride, it's not going to dissolve anymore because that solution has taken as much as it can take, and therefore what will happen is you're going to have undissolved sodium chloride at the bottom. So we're talking about a saturated solution, a solution that can take as much sodium chloride as it can take, and now it's beyond the point where you can't dissolve anymore sodium chloride. At that point, you'll have solid material at the bottom, all right? Now when you have solid material at the bottom, what happens is that you will establish an equilibrium where you have Na plus equals plus Cl. In other words, you will establish an equilibrium where when one unit of NaCl dissolves, so one cation and anion dissolves, somewhere else they will deposit. One unit will deposit. You see that? So if the solution is saturated, then what happens is you will establish this equilibrium where you have the solid material dissolving, and you'll have, if the solid material dissolves at any point, you're going to have two dissociated cations and anions associating with each other and depositing on top of that solid, okay? And sometimes the solid material at the bottom is called, we call this a precipitate. So when you have undissolved material at the bottom, we say we have a precipitate at the bottom, okay? Now since this is an equilibrium, we should be able to apply the law of mass action to this scenario, but it turns out we can't because this solution is non-ideal. This solution is non-ideal, all right? And the reason it's non-ideal, the solution is non-ideal because it is not homogeneous. And the reason it's not homogeneous is that when the solution is saturated, the saturated solution has mixtures of ion pairs and ion clusters, all right? So the reason that the solution is non-ideal is because it's so saturated, now the cations and anions are so crowded by chance that they're so close to each other that they pair up. So we call that an ion pair. So when the solution is so crowded with cations and anions, now in solution you'll have one cation and anion so close to each other that they pair up and you have an ion pair. Or by pure chance if four of them are together, two pairs of cations and anions will form a cluster. You can have three pairs of cations and anions forming a three-pair cluster. You can have four pairs forming a four-pair cluster. So what happens is now you have all these different species in there. You just don't have cations and anions separated from each other. On top of that you have a pair paired up called an ion pair. You can have a few cations and anions forming clusters and the number of cations and anions in the cluster can vary. So the net result is now you have mixtures of different kinds of pairs and so we say the solution is non-ideal. Your solution has to be uniform. Your solution, if you're going to calculate concentrations and you want to determine how much of each species is there, if you have mixtures of all of this, you can have independent separate cations and anions. You can have in that solution not only that, but you can have pairs of ions. You can have one pair or you can have two, three, four, five pairs of ions clustered together and as a result the solution is no longer homogeneous. And because it's not homogeneous, we say it's a non-ideal solution. It's not an ideal solution is where everything is homogeneous, dissolved completely and all the species are the same. So ideal solution would be just having Na plus and Cl minus and that's it. But if you have clusters and ion pairs and so on, now it becomes non-ideal. So the net result is that the law of mass action cannot be applied. So when you take a salt that dissolves easily in water, a soluble salt, you can't solve an equilibria problem. You can't use the law of mass action because that solution is not ideal. But if you take insoluble, so we're going to call these insoluble salts, or we call them marginally soluble salts. So incompletely soluble salts, marginally soluble salts. There's another word that we use and I want and slightly soluble salts. So if you say insoluble salts, marginally soluble salts or slightly soluble salts. Now if you take these, they all mean the same thing. So sometimes we use the word insoluble, sometimes we use the word marginally soluble, sometimes we use the word slightly soluble. So these are the exact reverse soluble salts. All the salts are the ones that dissolve completely. Everything else falls under this category. And an example of this would be something like silver chloride. So silver chloride is a precipitate. It's insoluble and the reason it's insoluble is if you place it in water, it just sinks to the bottom. It just forms a precipitate. So now we don't need to form a saturated solution. The instant you put silver chloride down here, it will form a solid. It will form a precipitate. And because it's insoluble or marginally soluble or slightly soluble, a tiny, tiny, tiny amount actually dissociates. So it's like, this is the equivalent of like weak acids. Remember weak acids, weak bases dissociate only a small amount. The majority of it stays undissociated. The same thing here. So when you take an insoluble, marginally soluble, a partially soluble salt, now you set up this equilibrium. Now this is an ideal solution. Why? It's not saturated. This is a solution where only a very, very small amount of the silver and chloride dissolve. So in solution, they're far apart from each other. And so there's no chance that you can form ion pairs or ion clusters, okay? Because the amounts are so small and they're separated by billions and billions of water molecules that they don't form ion pairs or ion clusters. And so what you do is you establish this equilibrium where some solid will dissolve and then some of the dissolved material will combine to give you a solid and you would establish an equilibrium. And because this is an ideal solution now, we can apply the law of mass action. So since this is an ideal solution, the law of mass action can be applied. All right? So now only in the case of insoluble or marginally soluble or slightly soluble substances, we can apply the law of mass action. And if we can apply the law of mass action, then we can write the equilibrium we're going to look at, which is the solid precipitate in equilibrium with dissociated silver ions and chloride ions. Okay? Now can everybody see the difference between soluble salts and insoluble salts are their attraction to water. Soluble salts, the instant you put them in water, they dissociate to give you cations and anions because their attraction for water is greater than their attraction for each other. Insoluble, a margin of soluble salts, their attraction for each other is much greater than their attraction for water and that's why they form a precipitate. They don't dissociate in water, they'd rather stay associated as the solid. Okay? So if you look at silver chloride now, we can write, this is an equilibrium, but can everybody see that this equilibrium is different from the other equilibrium we looked at because we have a solid and we have the others dissolved in aqueous solution. When you have two different phases, what do we call that? Heterogeneous. And do you guys remember last quarter when we looked at the basic fundamental concepts of equilibria, when you deal with heterogeneous systems that have pure solids, pure liquids, what do we do? We don't include that in the law of mass action of the equilibrium expression. So in this case, if I were to write the equilibrium expression, which is the law of mass action, we'd say K and this has a special name, so we call the Ksp equals the silver ion concentration times the chloride ion concentration and this has a special name, this is called the solubility product constant. And remember the word solubility is the amount that dissolves, alright? And so you can see product means time, so what we're looking at is the solubility of silver and chloride, the amounts of silver and chloride that dissolve and you're multiplying that and that's what gives you the solubility product. See this is the solubility of AG plus, this is the solubility of chloride, this is the product of that and therefore we call it the solubility product constant. Because these are all substances that don't dissolve easily in water, would you expect Ksp to be a big number or a small number? Small. Small, because the equilibrium lies entirely on the solid side, only very, very minuscule amount actually will dissociate. So all Ksp values are really, really small. Remember the Ksp value is large, it'd be like sodium chloride, alright? If Ksp is really large, but then we can't apply Ksp to soluble substances because the solutions are non-ideal, alright? So we don't apply Ksp to soluble salts, we only apply them to insoluble, marginally soluble or slightly soluble, okay? So if you look at the value for this, this is about 1.6 times 10 to the negative 10. If you were to write the unit for this, you can see this is moles per liter, this will be moles per liter, so it will be moles squared per liter squared, okay? Often we, even though we know that the equilibrium constant has a unit, very often we omit that unit, okay? So you should remember that if it had a unit, this would be moles squared per liter squared, okay? And this would be at room temperature, alright? Now if I were to raise, remember we said that the equilibrium constant always depends on temperature. If you change the temperature, the value of this would change. And so if I raise the temperature to 100 degrees Celsius, now this value would be 2.2 times 10 to the negative 8. Now look at that and tell me, now does Ksp, as you raise the temperature, does Ksp increase or decrease? Increases. Remember, what does Ksp tell you? It's solubility product. It tells you what the solubility is. Whereas you raise the temperature, what happens to solubility? It increases. That means more dissolves. And we know that. If you want to dissolve a substance, you want to increase the solubility, what do you do? You heat it up, alright? And you can see when you heat up and increase the temperature, Ksp actually goes up. And if Ksp goes up, it tells us that the solubility has increased, okay? Now if you wanted to know, so let's take some other examples. So if I take another example, would be PbI2. This is an insoluble substance. The equilibrium you would set up would be Pb2 plus equus plus 2i minus equus. There are two i-bins there. So if you look at this formula, when it dissociates, it's going to give you 2i minus. However, in all of these equations, you have to maintain charge balance. So you can see we start with something that's electrically neutral. Therefore, this side has to be electrically neutral as well. Lead is 2 plus, and therefore it's got to have 2i minus as to balance that. And overall, 2 plus, 2 minus, they balance each other. And if you take the Ksp for this, this is of the order of 7.1 times 10 to the negative 9. If you take something like, and so if I want to write the law of mass action for this, this would be the concentration of Pb2 plus times the concentration of i taken to the power 2. If you take something like aluminum hydroxide, this is also an insoluble substance, the equilibrium you would establish is now, the way to figure out charge balance is to look at hydroxide. What is the charge on hydroxide minus 1? How many hydroxide are there? Three. So that means you have three minuses. To counterbalance that charge, you have only one aluminum. So what should the charge of aluminum come out to be? Three plus. So you have to figure out what the charge is. In every case, you have to figure out what the charge of your cation is and what the charge of the anion is. Now usually group one metals have a charge of what? Plus one. What is the charge of group two metals? Plus two. The transition metals are what? Variable. Variable. It changes. And so for transition metals, you don't know what it is. And if you want to figure it out, you always turn to the anion. So if you know the charge of the anion, you can figure out what the charge of the cation is. So in this case, aluminum comes out to be three plus and hydroxide. You need three hydroxides to balance that charge. And if you look at Ksp, Ksp comes out to be two times ten to the negative thirty-two. That's a tremendously small number. So you can see only a very, very tiny amount of that actually is soluble. The solubility is really, really small. And so if you want to write Ksp for that, Ksp would be the concentration of the aluminum cations times the concentration of the hydroxide ions taken to the power three. So because Ksp deals with substances that are insoluble, we know that all Ksp values have to be really small. You can never have a Ksp number that is greater than one, all right? Because that means there's a soluble salt and you can't have a Ksp for soluble salts, okay? So here is, I pulled this out of a handbook and it's really, really small and that's okay because what I really wanted you to focus on is that just like we have tables for Ka values, there are tables for Ksp values. And usually these are in alphabetical order, all right? And so you can see aluminum hydroxide and you can see the Ksp that corresponds to that. But what I wanted you to take a look at is, take a look at all the Ksp values. Can you see that all of these are numbers that are less than one because they're all insoluble, marginally soluble, or slightly soluble and they're all going to have really, really small numbers. Got it? So that's important to remember. Another thing that I want you to remember is that one way for you to identify whether a substance is insoluble or soluble is if it has a Ksp value that's really, really small, if it has a Ksp value at all, it has to be insoluble, marginally soluble, or slightly soluble. Another way to figure out whether a substance assault is soluble or not is to look at the solubility rules. In high school you have seen this before and we call these the solubility rules and this is a guide for you to figure out whether an ionic compound or salt will dissolve in water or not. So another way for you, rather than looking at Ksp values, another way to figure out whether an ionic compound or salt is soluble or insoluble, marginally soluble or slightly soluble is to follow these rules. Now I don't want you to memorize these rules but I do want you to be able to understand these rules and if you have the rules right next to you be able to apply them. So on an exam if I want you to use this I will actually provide the rules. I'm more interested in whether you can actually apply the rules rather than whether you are good at memorizing the rules. So the first rule is that all nitrates and acetates are soluble. So if you take an ionic compound where the anion is a nitrate or the anion is an acetate, a sodium acetate, potassium acetate, all of these will dissolve completely. So the first rule is that nitrates, ionic compounds with anions that are nitrates or acetates are soluble in water. Sometimes we write the formula for acetate like that, sometimes we write it as CH3CO2- they're the same thing. So CH3CO2- is what? The conjugate base of the weak acid is acidic acid. So acetate is the conjugate base of the weak acid acidic acid. The second rule deals with chlorides, bromides and iodides and what we know is all chlorides are soluble except when the cation is silver, when the cation is mercury and when the cation is lead. So except for those all chlorides are soluble. The same rule applies for bromides and the same rule applies for iodides. The third rule is all sulfates are soluble. So if your anion is a sulfate, all sulfates are soluble except barium, strontium and lead. They're all insoluble. Then they say calcium, mercury and silver are only slightly soluble. Remember for all intents and purposes we're going to say that insoluble mean the same thing as slightly soluble mean the same thing as marginally soluble. It's like categorizing them all as weak acids. Either they're weak acids or strong acids. In this case they're either they dissolve completely which are the soluble salts or the opposite. Yes, go ahead. What's the S after the ZA? That's a typo. Thanks for pointing that out. It's calcium sulfate. That's sulfur. There are two S's there and can you cross out the lower case S? So calcium sulfate, mercury sulfate, mercury sulfate and silver sulfate are only slightly soluble. The third one is all group 1 metal ions and ammonium ions are soluble. So if you have sodium potassium or if your cation is ammonium they're soluble in water. All hydroxides are insoluble except group 1 metal ions and ammonium ions. Barium, strontium, calcium, these are group 2 metals. Barium, strontium, calcium are marginally soluble. The set of rules are on the class website so you can print it out. So I put it in, if you look at yesterday's date I put this up already so you can print it out. Six is all sulfides are soluble except group 1 metal ions and ammonium. Calcium ions, strontium ions and barium ions are also insoluble. All phosphates and carbonates are insoluble except group 1 metal ions and ammonium ions. Most chromates are only slightly soluble except those are group 1 metal ions and ammonium ions. Okay? So this covers almost every anion or cation combination that you can come across. All right? Now what I want to keep in mind is there are some very rare instances where you may have two rules that give you kind of, it's very, very rare, that two rules will give you opposite information. All right? If that does happen then what you have to make keep in mind is any rule on top has precedence over any rule at the bottom. Is that clear? It kind of goes in that order and if you encounter it's very, very rare. It's very unlikely you're going to encounter it but if you ever encounter where two rules kind of give you conflicting information then always make sure that anything on top takes precedence over anything below. Got it? So either you can use KSP values or you can use solubility rules to figure out whether anionic salt is soluble or not. Okay? So now we can, now that we understand the difference between soluble salts and insoluble salts and that why we focus only on the insoluble or the marginally soluble or the slightly soluble compounds now we're ready to look at solubility and KSP. All right? So we want to look at the relationship between solubility and KSP. We said before if the KSP gets bigger or any substance that has a KSP that has a bigger number we know that's related to solubility so that means that substance will be dissolving a little bit more than something that has a lower KSP. Okay? So solubility refers to the amount of the substance that dissolves. Okay? So if we want to look at solubility and let's take an example of silver chloride. So we know silver chloride dissociates to give you silver, cations and chloride anions and we know this equilibrium is given by KSP and the equilibrium constant is a really, really small number. Now we want to calculate what is the solubility? How much of this actually dissolves? All right? To do that we don't need to go through the complete ice chart because you look at, you usually use the ice chart where you look at the initial, then the change and then at equilibrium if you need the initial concentration. Now if you look at this, am I interested in the initial concentration of silver chloride? It doesn't dissolve in water anyway. All right? It doesn't appear in the law of mass action. So all I'm interested in looking at is what do I have at equilibrium? Can everybody see at equilibrium if I look at the concentrations? We know that if negative x dissociates we know plus x plus x would form. So at equilibrium all I'm left with is just that, xx. Can everybody see that? So all I'm interested is the amount of silver and chloride at equilibrium and since they dissociate in a one to one stoichiometry that's going to be x. So I know that Ksp equals 1.6 times 10 to the negative 10 which gives me the silver ion concentration times the chloride ion concentration which is x times x which is x squared. Okay? So if I take the square root of that I end up with 1.3 times 10 to the negative 5 moles per liter. Now this is the amount that dissolved. Remember this is the amount that's consumed. This is negative x and the amount that's so. Since this is in moles per liter we call this molar solubility. All right? So we have in general solubility and this gives the amount or concentration of AgCl that dissolved. All right? Remember solubility is related to how much actually dissolves, how much is soluble and therefore when you talk about molar solubility it gives us in terms of concentration the amount of the starting material that dissolved which is x. Now sometimes it's customary to give this as gram solubility. So if you want to give this in terms of gram solubility so here this is molar solubility and if it's molar solubility the unit is moles per liter. All right? If you give it in grams solubility you want to know how many grams of the solid material actually dissolved. All right? And so if I want to convert moles to grams what do I need to do? I need to multiply by the molar mass. All right? So I'm going to start with 1.3 times the n to the negative 5 moles per liter and I'm going to multiply by the molar mass of silver chloride. So you go look at the molar mass of silver, the molar mass of chloride, add them together that gives me 143.3 grams of silver chloride per mole. So this is the molar mass, this is times the molar mass of AgCl which is the reactant because solubility refers to the amount of the reactant that dissolved. Okay? So you can see moles and moles will cancel out and so I end up with 1.9 times 10 to the negative 3 grams per mole of AgCl. So you can calculate solubility in terms of concentration and that is x, that's the amount that dissolved, that would be molar solubility or you can calculate in terms of grams and that would be gram solubility. Oh I'm sorry grams per liter, yes. Alright so the unit would be grams per liter, grams per mole would be the molar mass. Got it? So let's apply this in a problem. So let's take a very straightforward and easy problem and so if we take this example where the solubility product constant for calcium fluoride is given to you. Okay? So you calculate the concentrations of calcium and fluoride in a saturated solution of calcium fluoride at 25 degrees Celsius and determine the solubility of calcium fluoride in grams per liter. Now here they're using the word saturated and when you talk about insoluble substance the saturated refers to just the teeny tiny amount that will dissolve. Alright? When you talk about soluble salts if you use the word saturated it means that it's super super saturated. Do you get it? So that's the difference. So you should be able to recognize from the KSP that's 10 to the negative 11 and so this is an insoluble substance so very little actually dissolves and when the maximum amount that can dissolve is dissolved we call that a saturated solution but the maximum amount that an insoluble salt dissolves is going to be really a minuscule amount. Got it? So we're looking at calcium fluoride so we're going to set up the equilibrium so we know this is insoluble and we're looking at KSP so the equilibrium that you're going to establish would be this. Calcium is group 2 so it's got to have 2 plus, fluoride is group 7, it's got to have minus 1 charge, there are 2 of those and we know KSP equals 3.9 times 10 to the negative 11 that number is going to be really really small. That means that if I want to calculate, I know KSP, if I want to calculate the concentration of calcium and fluoride I know at equilibrium if X is consumed how much calcium would form X but if X is consumed how much fluoride would form to X. Okay so don't forget that and therefore we know that KSP equals calcium ion concentration times fluoride ion concentration raised to the power 2 which is X times 2X squared. Make sure you put this within parenthesis because if you take 2X squared that becomes 4X squared times X, what would that give me? 4X cubed. So we know that 4X cubed should equal 3.9 times 10 to the negative 11 which is KSP therefore if you work out the value of X it comes out to be 2.1 times 10 to the negative 4 moles per liter. So X is solubility, this is molar solubility, this is the amount of solid that dissolved. The first part is they want us to calculate the calcium ion concentration so therefore the calcium ion concentration is actually X and that would be 2.1 times 10 to the negative 4 molar. If you take the fluoride ion concentration the fluoride ion concentration would be what? 2X and therefore it should be 4.2 times 10 to the negative 4 molar. Now lastly they want you to calculate the solubility of calcium fluoride in grams per liter. So now we want to calculate gram solubility and that would be the molar solubility which is 2.1 times 10 to the negative 4 moles per liter times the molar mass of calcium fluoride. So look at the periodic chart, calculate the molar mass of calcium plus 2 fluorides and if you add them up the number comes out to be 78.1 grams per mole, moles and moles are cancelled out and this is calcium fluoride. And therefore what you end up with is 0.016 grams per liter of calcium fluoride. So the gram solubility gives you the mass of calcium fluoride that actually dissolved per liter and that we figured out. So next class we're going to take a few more examples of solving equilibrium problems associated with KSP. So we'll stop there for today.
 A quick announcement.  I wanted to remind you that yesterday you would have received a PDF file of your graded  exam.  So take a look at the exam and if you have any questions about the grading, you should  print out the entire exam and take it with you to discussion.  And talk to your TA about it.  If you do think there is an error, then what you need to do is print out a copy of the  exam, attach a note explaining to me what the problem is, and you need to hand it to  me, not the TAs.  So all regrades have to be turned into me.  Please do not send me an email because as you know, when it's a class of this size, it's  just I can't be non-dated with PDF files and it's difficult for me to keep track of who  sent what.  So that's why I'm asking you to print it out, attach a note, and then turn it into me,  and the deadline is two weeks from yesterday because it's two weeks from the date when  you received the PDF files from me.  And so within that deadline, if you turn it in, then we will take a look at it.  Is that clear to everybody?  So please do not send me an email.  You can hand it to me after class or you can hand it to me before class.  You can hand it to me during my office hours.  And if I'm not there, you can even slide it out of my door as long as I get it within  the two week deadline.  Is that clear to everybody?  So if you guys remember, we're looking at acid-base titrations and we said we're going  to look at three types of acid-base titrations.  Strong acid, strong base of isewersa.  We looked at weak acid, strong base.  Now we're going to look at the third scenario where you have a weak base versus a strong  acid.  So if you look at the titration, you can see that the weak base is placed in the Erlenmey  flask.  And that is typically in an analysis, that is the one that has the unknown concentration.  We placed the strong acid in the burette.  And when you start the titration now, we want to look at what the reaction would be.  Remember we said that if you have an acid in a base, they can't coexist.  Large amounts of acids and bases do not coexist.  They will react to give you water and salt.  So let's see what happens here.  So if you write the complete ionic equation, and we'll start with the strong acid.  A strong acid we know dissociates completely.  So in reality when it's dissolved in water, you have hydronium ion and chloride ion.  So that's the strong acid.  Now we have the weak base.  We know weak bases do not dissociate, do not react with water.  They tend to be undissociated.  Or in this case, they would not pick up a proton because it's a weak base.  We know that when we see the word weak, it means its preference is to remain the way  it is.  So this is what we have at the start.  Now we're going to let it go to products.  Now this is the base.  This is the acid.  We know they're going to react to give me NH4 plus aqueous plus water plus Cl minus aqueous.  And so what you end up with is this strong acid and the weak base will give you its conjugate  acid and water.  And if you look at this, chloride ions are there at the beginning, chloride ions are  there at the end.  So the chloride ions are the spectator ions.  So we have only one spectator ion here, which is the chloride ion.  Now if you write the net ionic equation, now we take away the spectator ions.  So because remember the spectator ions don't participate in the reaction, they're just  there.  And so if I write the net ionic equation, what I end up with is the hydronium ion plus  ammonia, which is the acid and the base, will react to give me water NH4 plus, which is  the conjugate acid and water.  So this is the acid.  This is the base.  The acid will donate a proton to the base so that you end up with NH4 plus and water.  So this is the neutralization.  And so if we look at this, if all of this reacted completely with this at the equivalence  point or the stoichiometric point, what you'd have is water.  Water is neutral plus you have NH4 plus.  So now look at NH4 plus.  NH4 plus is the conjugate acid of a weak base, which is NH3.  So if you take NH4 plus, would you expect it to be acidic, basic, or neutral?  Acidic.  All right?  At the equivalence point now, the pH has to be acidic.  All right?  And then if you write the overall reaction or the general form of the reaction, it will  be where you have HCl, which is the strong acid reacting with ammonia, which is the weak  base giving you NH4 plus.  In fact, I'm writing the general form of the reaction.  So it just turned out to be giving you salt here.  All right?  So if you write it in the general form, you see that what you're forming is ammonium chloride.  You're forming the salt, the strong acid reacting with the base.  So can everybody see that the general form or the overall reaction or sometimes they  call this the molecular equation, all right, where it identifies the molecules?  See that these are covalent compounds, so you can use the word molecular, all right?  And this is also called the molecular equation.  Okay?  So now let's look at what the pH titration curve looks like.  So this is what the titration curve looks like.  So over here, we're measuring the pH on this side.  On the y-axis, we have the pH of the analyte.  Over here, we have the volume of the strong acid that's added.  And you can see now the shape of this is very different.  Okay?  To begin with, you just have a weak base, all right?  Before we add any strong acid, we have only a weak base in the flat.  So we'd expect the pH to be high, all right?  It's a basic solution.  Now as you add a strong acid and you're neutralizing the weak base and converting it to NH4 plus,  which is a weak acid, you know that the pH is going to go down.  Okay?  So what we see is that the pH decreases.  But in this region, this is the buffer region, all right?  Once again, over here in this region here, it would act as a buffer.  So that's why it flattens out.  The pH doesn't change dramatically there.  It just kind of flattens out in plateaus because now what we have is in that mix, in that solution,  you'll have NH3, which is the weak base, and you're going to have NH4 plus, which is its  conjugate acid, and you have mixtures of both in substantial amounts.  And so in that region, it's going to act as a buffer, all right?  So that's where this flattens out.  And then as you add more and more acid, now your buffer is going to get completely swamped  and everything is going to get converted to NH4 plus.  So at the equivalence point, now you can see all I have is NH4 plus, Cl minus, and water.  Cl minus is the conjugate base of a strong acid.  We know that's neutral.  So Cl minus is neutral.  Water is neutral, all right?  And therefore, the only other species you have in there is NH4 plus, and the NH4 plus  is the conjugate acid of a weak base.  It's going to be acidic.  And therefore, the pH is less than 7, all right?  So you can see that the pH is less than 7, okay?  Beyond this point, now you're swamped with a strong acid.  So beyond the equivalence point, the pH is going to determine by the extra hydrochloric  acid that's there, which associates completely.  And you can see beyond the equivalence point, now rapidly the solution gets very acidic,  all right?  Now if this is the equivalence point, so you can see that in this titration, the concentrations  are different, okay?  In this titration, it turns out that the equivalence point is at 50 milliliters, all right?  So half the equivalent point would be 25.  So if you take 25 milliliters, that would correspond to that.  And so at 25 milliliters, we know that we have a buffer there where the concentrations  of the weak acid are equal to each other.  So the concentrations are equal to each other, you know, it's a buffer.  And if the concentrations are equal to each other, the ratio of acid to base would be  1.  Log 1 would be what?  0.  And therefore, we know pH equals pKa, all right?  So here, even though we have a base and solution, this gives me the point at which pH equals  pKa, all right?  It's not pH equals pKa, even though we have an acid down there, some students say, well,  we have a base down there, so why are we talking about kA rather than kB?  But remember, pH always equals pKa, according to Henderson-Husselbalck equation, where the  concentrations of acid and base are the same, all right?  So here, this would be pH equals pKa.  And so you don't even have to do any calculations.  I ask you, what is the pH of the half-equivalence point?  You just look at this table and you look for the conjugate acid-base pair that we're interested  in looking at, and that would be ammonium and ammonia.  And you know, pKa is 9.25, so what should this pH correspond to?  9.25, all right?  Because we know that at the half-equivalence point, the pH has to equal pKa.  Is that clear to everybody?  So now we've looked at all three types of titrations.  I did not do any calculations here, but you should be able to do the same type of calculations  that we did for weak acid-strong base or strong acid-strong base.  And I'd like you to try that at home.  In fact, I've assigned a problem from your textbook where you get to practice that.  Okay?  Yes, go ahead.  Can you explain what happened to the equivalent point, meaning the volume?  So her question was, what happens at the equivalent point?  So you have to turn to the net ionic equation.  So in this titration, what's happening is the strong acid neutralizes the weak base,  all right?  And so essentially what happens is, when this reaction takes place, you have NH4 plus and  water being formed.  Okay?  So when you have stoichiometric amounts of both or equivalent amounts of both, where  all of this reacts completely with this and all of this reacts completely with this, so  that you have no excess reagent left over, all right?  Both reactants react completely.  And that happens only at the equivalent point because you have to have stoichiometric amounts.  You have to have the right amount of each so that you don't have any limiting reagent.  You don't have any excess reagent.  They react in stoichiometric amounts.  So that only happens at the equivalent point, okay?  So at the equivalent point, when both of the reactants have reacted completely, all that  you have left behind is just the product.  You don't have any excess reagent left over.  So because all of them are reacted, all that you have there is just NH4 plus, water, and  Cl minus, all right?  The spectator ion is Cl minus.  Cl minus is the conjugate base of a strong acid, and we know that that's neutral.  We know water is neutral, all right?  And so now we turn to NH4 plus.  So those are only three species in solution.  If you turn to NH4 plus, you know that that's a weak acid.  It's the conjugate acid of a weak base, and therefore it's a weak acid.  If it's a weak acid, it's going to be acidic, all right?  And therefore the pH should be less than 7.  Got it?  Okay, so now we've completed looking at acid-based titrations, and now we're actually ready to  move on to the last topic that we're going to look at in terms of titrations.  So by now, you guys are experts at solving any equilibrium problem.  And we focused, this entire quarter, we focused on equisiculabria, what happens in water, all  right?  And that kind of has been the focus all along.  Now we're going to look at the last type of equilibria, and this deals with solubility,  all right?  So we're going to look at solubility product.  And as all of you know, solubility deals with when a substance dissolves.  So we're looking at solubility product in equisolutions, and so the word solubility is  related to soluble when something dissolves, all right?  So let's take something, as all of you know, we know that many, many ionic compounds dissolve  easily in water, and we call them soluble salts.  So if I take soluble salts.  So we'll start by looking at soluble salts.  And all of you know, when we use the word salts, we're talking about ionic compounds,  and we said if you take ionic compounds that are soluble, what happens the instant you  dissolve them in water, what happens to soluble salts?  They dissolve, and when they dissolve, what does that tell you?  They dissociate completely, okay?  So give me an example of a really good soluble salt in ACL.  We all know table salt in ACL is a very good soluble salt.  So I start with an ACL, and all of you know that if I take a beaker, and I'm going to  add solid sodium chloride.  So you take, all of you know that ionic compounds tend to be crystalline solids, and if you  take common table salt, you can see that they have nice crystalline granules.  And so if you take a spoonful of that and put it in a beaker water and stir it, what's  going to happen?  It dissolves completely.  And when we say it dissolves completely, we know that it has dissociated completely  to give you Na plus and Cl minus, okay?  Now what I'm going to do is I'm going to add another spoonful and stir it up.  It's going to dissolve.  I'm going to take another spoonful and stir it up.  It'll dissolve.  I'm going to keep doing this until the point comes when that solution is completely saturated  with sodium chloride.  So if I keep putting more and more and more sodium chloride, what I'll end up with is  a really saturated solution.  And the point at which this gets to be saturated, we know it's saturated because it can't take  anymore.  If you add another spoonful of sodium chloride, it's not going to dissolve anymore because  that solution has taken as much as it can take, and therefore what will happen is you're  going to have undissolved sodium chloride at the bottom.  So we're talking about a saturated solution, a solution that can take as much sodium chloride  as it can take, and now it's beyond the point where you can't dissolve anymore sodium chloride.  At that point, you'll have solid material at the bottom, all right?  Now when you have solid material at the bottom, what happens is that you will establish an  equilibrium where you have Na plus equals plus Cl.  In other words, you will establish an equilibrium where when one unit of NaCl dissolves, so one  cation and anion dissolves, somewhere else they will deposit.  One unit will deposit.  You see that?  So if the solution is saturated, then what happens is you will establish this equilibrium  where you have the solid material dissolving, and you'll have, if the solid material dissolves  at any point, you're going to have two dissociated cations and anions associating with each other  and depositing on top of that solid, okay?  And sometimes the solid material at the bottom is called, we call this a precipitate.  So when you have undissolved material at the bottom, we say we have a precipitate at the  bottom, okay?  Now since this is an equilibrium, we should be able to apply the law of mass action to  this scenario, but it turns out we can't because this solution is non-ideal.  This solution is non-ideal, all right?  And the reason it's non-ideal, the solution is non-ideal because it is not homogeneous.  And the reason it's not homogeneous is that when the solution is saturated, the saturated  solution has mixtures of ion pairs and ion clusters, all right?  So the reason that the solution is non-ideal is because it's so saturated, now the cations  and anions are so crowded by chance that they're so close to each other that they pair up.  So we call that an ion pair.  So when the solution is so crowded with cations and anions, now in solution you'll have one  cation and anion so close to each other that they pair up and you have an ion pair.  Or by pure chance if four of them are together, two pairs of cations and anions will form  a cluster.  You can have three pairs of cations and anions forming a three-pair cluster.  You can have four pairs forming a four-pair cluster.  So what happens is now you have all these different species in there.  You just don't have cations and anions separated from each other.  On top of that you have a pair paired up called an ion pair.  You can have a few cations and anions forming clusters and the number of cations and anions  in the cluster can vary.  So the net result is now you have mixtures of different kinds of pairs and so we say  the solution is non-ideal.  Your solution has to be uniform.  Your solution, if you're going to calculate concentrations and you want to determine how  much of each species is there, if you have mixtures of all of this, you can have independent  separate cations and anions.  You can have in that solution not only that, but you can have pairs of ions.  You can have one pair or you can have two, three, four, five pairs of ions clustered together  and as a result the solution is no longer homogeneous.  And because it's not homogeneous, we say it's a non-ideal solution.  It's not an ideal solution is where everything is homogeneous, dissolved completely and all  the species are the same.  So ideal solution would be just having Na plus and Cl minus and that's it.  But if you have clusters and ion pairs and so on, now it becomes non-ideal.  So the net result is that the law of mass action cannot be applied.  So when you take a salt that dissolves easily in water, a soluble salt, you can't solve  an equilibria problem.  You can't use the law of mass action because that solution is not ideal.  But if you take insoluble, so we're going to call these insoluble salts, or we call  them marginally soluble salts.  So incompletely soluble salts, marginally soluble salts.  There's another word that we use and I want  and slightly soluble salts.  So if you say insoluble salts, marginally soluble salts or slightly soluble salts.  Now if you take these, they all mean the same thing.  So sometimes we use the word insoluble, sometimes we use the word marginally soluble,  sometimes we use the word slightly soluble.  So these are the exact reverse soluble salts.  All the salts are the ones that dissolve completely.  Everything else falls under this category.  And an example of this would be something like silver chloride.  So silver chloride is a precipitate.  It's insoluble and the reason it's insoluble is if you place it in water, it just sinks  to the bottom.  It just forms a precipitate.  So now we don't need to form a saturated solution.  The instant you put silver chloride down here, it will form a solid.  It will form a precipitate.  And because it's insoluble or marginally soluble or slightly soluble, a tiny, tiny, tiny amount  actually dissociates.  So it's like, this is the equivalent of like weak acids.  Remember weak acids, weak bases dissociate only a small amount.  The majority of it stays undissociated.  The same thing here.  So when you take an insoluble, marginally soluble, a partially soluble salt, now you  set up this equilibrium.  Now this is an ideal solution.  Why?  It's not saturated.  This is a solution where only a very, very small amount of the silver and chloride dissolve.  So in solution, they're far apart from each other.  And so there's no chance that you can form ion pairs or ion clusters, okay?  Because the amounts are so small and they're separated by billions and billions of water  molecules that they don't form ion pairs or ion clusters.  And so what you do is you establish this equilibrium where some solid will dissolve and then some  of the dissolved material will combine to give you a solid and you would establish an equilibrium.  And because this is an ideal solution now, we can apply the law of mass action.  So since this is an ideal solution, the law of mass action can be applied.  All right?  So now only in the case of insoluble or marginally soluble or slightly soluble substances, we  can apply the law of mass action.  And if we can apply the law of mass action, then we can write the equilibrium we're going  to look at, which is the solid precipitate in equilibrium with dissociated silver ions  and chloride ions.  Okay?  Now can everybody see the difference between soluble salts and insoluble salts are their  attraction to water.  Soluble salts, the instant you put them in water, they dissociate to give you cations  and anions because their attraction for water is greater than their attraction for each  other.  Insoluble, a margin of soluble salts, their attraction for each other is much greater  than their attraction for water and that's why they form a precipitate.  They don't dissociate in water, they'd rather stay associated as the solid.  Okay?  So if you look at silver chloride now, we can write, this is an equilibrium, but can  everybody see that this equilibrium is different from the other equilibrium we looked at because  we have a solid and we have the others dissolved in aqueous solution.  When you have two different phases, what do we call that?  Heterogeneous.  And do you guys remember last quarter when we looked at the basic fundamental concepts  of equilibria, when you deal with heterogeneous systems that have pure solids, pure liquids,  what do we do?  We don't include that in the law of mass action of the equilibrium expression.  So in this case, if I were to write the equilibrium expression, which is the law of mass action,  we'd say K and this has a special name, so we call the Ksp equals the silver ion concentration  times the chloride ion concentration and this has a special name, this is called the  solubility product constant.  And remember the word solubility is the amount that dissolves, alright?  And so you can see product means time, so what we're looking at is the solubility of  silver and chloride, the amounts of silver and chloride that dissolve and you're multiplying  that and that's what gives you the solubility product.  See this is the solubility of AG plus, this is the solubility of chloride, this is the  product of that and therefore we call it the solubility product constant.  Because these are all substances that don't dissolve easily in water, would you expect  Ksp to be a big number or a small number?  Small.  Small, because the equilibrium lies entirely on the solid side, only very, very minuscule  amount actually will dissociate.  So all Ksp values are really, really small.  Remember the Ksp value is large, it'd be like sodium chloride, alright?  If Ksp is really large, but then we can't apply Ksp to soluble substances because the  solutions are non-ideal, alright?  So we don't apply Ksp to soluble salts, we only apply them to insoluble, marginally soluble  or slightly soluble, okay?  So if you look at the value for this, this is about 1.6 times 10 to the negative 10.  If you were to write the unit for this, you can see this is moles per liter, this will  be moles per liter, so it will be moles squared per liter squared, okay?  Often we, even though we know that the equilibrium constant has a unit, very often we omit that  unit, okay?  So you should remember that if it had a unit, this would be moles squared per liter squared,  okay?  And this would be at room temperature, alright?  Now if I were to raise, remember we said that the equilibrium constant always depends on  temperature.  If you change the temperature, the value of this would change.  And so if I raise the temperature to 100 degrees Celsius, now this value would be 2.2 times  10 to the negative 8.  Now look at that and tell me, now does Ksp, as you raise the temperature, does Ksp increase  or decrease?  Increases.  Remember, what does Ksp tell you?  It's solubility product.  It tells you what the solubility is.  Whereas you raise the temperature, what happens to solubility?  It increases.  That means more dissolves.  And we know that.  If you want to dissolve a substance, you want to increase the solubility, what do you do?  You heat it up, alright?  And you can see when you heat up and increase the temperature, Ksp actually goes up.  And if Ksp goes up, it tells us that the solubility has increased, okay?  Now if you wanted to know, so let's take some other examples.  So if I take another example, would be PbI2.  This is an insoluble substance.  The equilibrium you would set up would be Pb2 plus equus plus 2i minus equus.  There are two i-bins there.  So if you look at this formula, when it dissociates, it's going to give you 2i minus.  However, in all of these equations, you have to maintain charge balance.  So you can see we start with something that's electrically neutral.  Therefore, this side has to be electrically neutral as well.  Lead is 2 plus, and therefore it's got to have 2i minus as to balance that.  And overall, 2 plus, 2 minus, they balance each other.  And if you take the Ksp for this, this is of the order of 7.1 times 10 to the negative  9.  If you take something like, and so if I want to write the law of mass action for this,  this would be the concentration of Pb2 plus times the concentration of i taken to the  power 2.  If you take something like aluminum hydroxide, this is also an insoluble substance, the equilibrium  you would establish is now, the way to figure out charge balance is to look at hydroxide.  What is the charge on hydroxide minus 1?  How many hydroxide are there?  Three.  So that means you have three minuses.  To counterbalance that charge, you have only one aluminum.  So what should the charge of aluminum come out to be?  Three plus.  So you have to figure out what the charge is.  In every case, you have to figure out what the charge of your cation is and what the  charge of the anion is.  Now usually group one metals have a charge of what?  Plus one.  What is the charge of group two metals?  Plus two.  The transition metals are what?  Variable.  Variable.  It changes.  And so for transition metals, you don't know what it is.  And if you want to figure it out, you always turn to the anion.  So if you know the charge of the anion, you can figure out what the charge of the cation  is.  So in this case, aluminum comes out to be three plus and hydroxide.  You need three hydroxides to balance that charge.  And if you look at Ksp, Ksp comes out to be two times ten to the negative thirty-two.  That's a tremendously small number.  So you can see only a very, very tiny amount of that actually is soluble.  The solubility is really, really small.  And so if you want to write Ksp for that, Ksp would be the concentration of the aluminum  cations times the concentration of the hydroxide ions taken to the power three.  So because Ksp deals with substances that are insoluble, we know that all Ksp values  have to be really small.  You can never have a Ksp number that is greater than one, all right?  Because that means there's a soluble salt and you can't have a Ksp for soluble salts,  okay?  So here is, I pulled this out of a handbook and it's really, really small and that's  okay because what I really wanted you to focus on is that just like we have tables for Ka  values, there are tables for Ksp values.  And usually these are in alphabetical order, all right?  And so you can see aluminum hydroxide and you can see the Ksp that corresponds to that.  But what I wanted you to take a look at is, take a look at all the Ksp values.  Can you see that all of these are numbers that are less than one because they're all insoluble,  marginally soluble, or slightly soluble and they're all going to have really, really small  numbers.  Got it?  So that's important to remember.  Another thing that I want you to remember is that one way for you to identify whether  a substance is insoluble or soluble is if it has a Ksp value that's really, really small,  if it has a Ksp value at all, it has to be insoluble, marginally soluble, or slightly  soluble.  Another way to figure out whether a substance assault is soluble or not is to look at the  solubility rules.  In high school you have seen this before and we call these the solubility rules and this  is a guide for you to figure out whether an ionic compound or salt will dissolve in water  or not.  So another way for you, rather than looking at Ksp values, another way to figure out whether  an ionic compound or salt is soluble or insoluble, marginally soluble or slightly soluble is to  follow these rules.  Now I don't want you to memorize these rules but I do want you to be able to understand  these rules and if you have the rules right next to you be able to apply them.  So on an exam if I want you to use this I will actually provide the rules.  I'm more interested in whether you can actually apply the rules rather than whether you are  good at memorizing the rules.  So the first rule is that all nitrates and acetates are soluble.  So if you take an ionic compound where the anion is a nitrate or the anion is an acetate,  a sodium acetate, potassium acetate, all of these will dissolve completely.  So the first rule is that nitrates, ionic compounds with anions that are nitrates or  acetates are soluble in water.  Sometimes we write the formula for acetate like that, sometimes we write it as CH3CO2-  they're the same thing.  So CH3CO2- is what?  The conjugate base of the weak acid is acidic acid.  So acetate is the conjugate base of the weak acid acidic acid.  The second rule deals with chlorides, bromides and iodides and what we know is all chlorides  are soluble except when the cation is silver, when the cation is mercury and when the cation  is lead.  So except for those all chlorides are soluble.  The same rule applies for bromides and the same rule applies for iodides.  The third rule is all sulfates are soluble.  So if your anion is a sulfate, all sulfates are soluble except barium, strontium and lead.  They're all insoluble.  Then they say calcium, mercury and silver are only slightly soluble.  Remember for all intents and purposes we're going to say that insoluble mean the same  thing as slightly soluble mean the same thing as marginally soluble.  It's like categorizing them all as weak acids.  Either they're weak acids or strong acids.  In this case they're either they dissolve completely which are the soluble salts or  the opposite.  Yes, go ahead.  What's the S after the ZA?  That's a typo.  Thanks for pointing that out.  It's calcium sulfate.  That's sulfur.  There are two S's there and can you cross out the lower case S?  So calcium sulfate, mercury sulfate, mercury sulfate and silver sulfate are only slightly  soluble.  The third one is all group 1 metal ions and ammonium ions are soluble.  So if you have sodium potassium or if your cation is ammonium they're soluble in water.  All hydroxides are insoluble except group 1 metal ions and ammonium ions.  Barium, strontium, calcium, these are group 2 metals.  Barium, strontium, calcium are marginally soluble.  The set of rules are on the class website so you can print it out.  So I put it in, if you look at yesterday's date I put this up already so you can print  it out.  Six is all sulfides are soluble except group 1 metal ions and ammonium.  Calcium ions, strontium ions and barium ions are also insoluble.  All phosphates and carbonates are insoluble except group 1 metal ions and ammonium ions.  Most chromates are only slightly soluble except those are group 1 metal ions and ammonium  ions.  Okay?  So this covers almost every anion or cation combination that you can come across.  All right?  Now what I want to keep in mind is there are some very rare instances where you may have  two rules that give you kind of, it's very, very rare, that two rules will give you opposite  information.  All right?  If that does happen then what you have to make keep in mind is any rule on top has precedence  over any rule at the bottom.  Is that clear?  It kind of goes in that order and if you encounter it's very, very rare.  It's very unlikely you're going to encounter it but if you ever encounter where two rules  kind of give you conflicting information then always make sure that anything on top takes  precedence over anything below.  Got it?  So either you can use KSP values or you can use solubility rules to figure out whether  anionic salt is soluble or not.  Okay?  So now we can, now that we understand the difference between soluble salts and insoluble  salts and that why we focus only on the insoluble or the marginally soluble or the slightly soluble  compounds now we're ready to look at solubility and KSP.  All right?  So we want to look at the relationship between solubility and KSP.  We said before if the KSP gets bigger or any substance that has a KSP that has a bigger  number we know that's related to solubility so that means that substance will be dissolving  a little bit more than something that has a lower KSP.  Okay?  So solubility refers to the amount of the substance that dissolves.  Okay?  So if we want to look at solubility and let's take an example of silver chloride.  So we know silver chloride dissociates to give you silver, cations and chloride anions  and we know this equilibrium is given by KSP and the equilibrium constant is a really,  really small number.  Now we want to calculate what is the solubility?  How much of this actually dissolves?  All right?  To do that we don't need to go through the complete ice chart because you look at, you  usually use the ice chart where you look at the initial, then the change and then at equilibrium  if you need the initial concentration.  Now if you look at this, am I interested in the initial concentration of silver chloride?  It doesn't dissolve in water anyway.  All right?  It doesn't appear in the law of mass action.  So all I'm interested in looking at is what do I have at equilibrium?  Can everybody see at equilibrium if I look at the concentrations?  We know that if negative x dissociates we know plus x plus x would form.  So at equilibrium all I'm left with is just that, xx.  Can everybody see that?  So all I'm interested is the amount of silver and chloride at equilibrium and since they  dissociate in a one to one stoichiometry that's going to be x.  So I know that Ksp equals 1.6 times 10 to the negative 10 which gives me the silver  ion concentration times the chloride ion concentration which is x times x which is  x squared.  Okay?  So if I take the square root of that I end up with 1.3 times 10 to the negative 5 moles  per liter.  Now this is the amount that dissolved.  Remember this is the amount that's consumed.  This is negative x and the amount that's so.  Since this is in moles per liter we call this molar solubility.  All right?  So we have in general solubility and this gives the amount or concentration of AgCl  that dissolved.  All right?  Remember solubility is related to how much actually dissolves, how much is soluble and  therefore when you talk about molar solubility it gives us in terms of concentration the  amount of the starting material that dissolved which is x.  Now sometimes it's customary to give this as gram solubility.  So if you want to give this in terms of gram solubility so here this is molar solubility  and if it's molar solubility the unit is moles per liter.  All right?  If you give it in grams solubility you want to know how many grams of the solid material  actually dissolved.  All right?  And so if I want to convert moles to grams what do I need to do?  I need to multiply by the molar mass.  All right?  So I'm going to start with 1.3 times the n to the negative 5 moles per liter and I'm  going to multiply by the molar mass of silver chloride.  So you go look at the molar mass of silver, the molar mass of chloride, add them together  that gives me 143.3 grams of silver chloride per mole.  So this is the molar mass, this is times the molar mass of AgCl which is the reactant  because solubility refers to the amount of the reactant that dissolved.  Okay?  So you can see moles and moles will cancel out and so I end up with 1.9 times 10 to the  negative 3 grams per mole of AgCl.  So you can calculate solubility in terms of concentration and that is x, that's the amount  that dissolved, that would be molar solubility or you can calculate in terms of grams and  that would be gram solubility.  Oh I'm sorry grams per liter, yes.  Alright so the unit would be grams per liter, grams per mole would be the molar mass.  Got it?  So let's apply this in a problem.  So let's take a very straightforward and easy problem and so if we take this example where  the solubility product constant for calcium fluoride is given to you.  Okay?  So you calculate the concentrations of calcium and fluoride in a saturated solution of calcium  fluoride at 25 degrees Celsius and determine the solubility of calcium fluoride in grams  per liter.  Now here they're using the word saturated and when you talk about insoluble substance  the saturated refers to just the teeny tiny amount that will dissolve.  Alright?  When you talk about soluble salts if you use the word saturated it means that it's super  super saturated.  Do you get it?  So that's the difference.  So you should be able to recognize from the KSP that's 10 to the negative 11 and so this  is an insoluble substance so very little actually dissolves and when the maximum amount that  can dissolve is dissolved we call that a saturated solution but the maximum amount that an insoluble  salt dissolves is going to be really a minuscule amount.  Got it?  So we're looking at calcium fluoride so we're going to set up the equilibrium so we know  this is insoluble and we're looking at KSP so the equilibrium that you're going to establish  would be this.  Calcium is group 2 so it's got to have 2 plus, fluoride is group 7, it's got to have  minus 1 charge, there are 2 of those and we know KSP equals 3.9 times 10 to the negative  11 that number is going to be really really small.  That means that if I want to calculate, I know KSP, if I want to calculate the concentration  of calcium and fluoride I know at equilibrium if X is consumed how much calcium would form  X but if X is consumed how much fluoride would form to X.  Okay so don't forget that and therefore we know that KSP equals calcium ion concentration  times fluoride ion concentration raised to the power 2 which is X times 2X squared.  Make sure you put this within parenthesis because if you take 2X squared that becomes  4X squared times X, what would that give me?  4X cubed.  So we know that 4X cubed should equal 3.9 times 10 to the negative 11 which is KSP therefore  if you work out the value of X it comes out to be 2.1 times 10 to the negative 4 moles  per liter.  So X is solubility, this is molar solubility, this is the amount of solid that dissolved.  The first part is they want us to calculate the calcium ion concentration so therefore  the calcium ion concentration is actually X and that would be 2.1 times 10 to the negative  4 molar.  If you take the fluoride ion concentration the fluoride ion concentration would be what?  2X and therefore it should be 4.2 times 10 to the negative 4 molar.  Now lastly they want you to calculate the solubility of calcium fluoride in grams per  liter.  So now we want to calculate gram solubility and that would be the molar solubility which  is 2.1 times 10 to the negative 4 moles per liter times the molar mass of calcium fluoride.  So look at the periodic chart, calculate the molar mass of calcium plus 2 fluorides and  if you add them up the number comes out to be 78.1 grams per mole, moles and moles are  cancelled out and this is calcium fluoride.  And therefore what you end up with is 0.016 grams per liter of calcium fluoride.  So the gram solubility gives you the mass of calcium fluoride that actually dissolved  per liter and that we figured out.  So next class we're going to take a few more examples of solving equilibrium problems associated  with KSP.  So we'll stop there for today.
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A quick announcement. I wanted to remind you that yesterday you would have received a PDF file of your graded exam. So take a look at the exam and if you have any questions about the grading, you should print out the entire exam and take it with you to discussion. And talk to your TA about it. If you do think there is an error, then what you need to do is print out a copy of the exam, attach a note explaining to me what the problem is, and you need to hand it to me, not the TAs. So all regrades have to be turned into me. Please do not send me an email because as you know, when it's a class of this size, it's just I can't be non-dated with PDF files and it's difficult for me to keep track of who sent what. So that's why I'm asking you to print it out, attach a note, and then turn it into me, and the deadline is two weeks from yesterday because it's two weeks from the date when you received the PDF files from me. And so within that deadline, if you turn it in, then we will take a look at it. Is that clear to everybody? So please do not send me an email. You can hand it to me after class or you can hand it to me before class. You can hand it to me during my office hours. And if I'm not there, you can even slide it out of my door as long as I get it within the two week deadline. Is that clear to everybody? So if you guys remember, we're looking at acid-base titrations and we said we're going to look at three types of acid-base titrations. Strong acid, strong base of isewersa. We looked at weak acid, strong base. Now we're going to look at the third scenario where you have a weak base versus a strong acid. So if you look at the titration, you can see that the weak base is placed in the Erlenmey flask. And that is typically in an analysis, that is the one that has the unknown concentration. We placed the strong acid in the burette. And when you start the titration now, we want to look at what the reaction would be. Remember we said that if you have an acid in a base, they can't coexist. Large amounts of acids and bases do not coexist. They will react to give you water and salt. So let's see what happens here. So if you write the complete ionic equation, and we'll start with the strong acid. A strong acid we know dissociates completely. So in reality when it's dissolved in water, you have hydronium ion and chloride ion. So that's the strong acid. Now we have the weak base. We know weak bases do not dissociate, do not react with water. They tend to be undissociated. Or in this case, they would not pick up a proton because it's a weak base. We know that when we see the word weak, it means its preference is to remain the way it is. So this is what we have at the start. Now we're going to let it go to products. Now this is the base. This is the acid. We know they're going to react to give me NH4 plus aqueous plus water plus Cl minus aqueous. And so what you end up with is this strong acid and the weak base will give you its conjugate acid and water. And if you look at this, chloride ions are there at the beginning, chloride ions are there at the end. So the chloride ions are the spectator ions. So we have only one spectator ion here, which is the chloride ion. Now if you write the net ionic equation, now we take away the spectator ions. So because remember the spectator ions don't participate in the reaction, they're just there. And so if I write the net ionic equation, what I end up with is the hydronium ion plus ammonia, which is the acid and the base, will react to give me water NH4 plus, which is the conjugate acid and water. So this is the acid. This is the base. The acid will donate a proton to the base so that you end up with NH4 plus and water. So this is the neutralization. And so if we look at this, if all of this reacted completely with this at the equivalence point or the stoichiometric point, what you'd have is water. Water is neutral plus you have NH4 plus. So now look at NH4 plus. NH4 plus is the conjugate acid of a weak base, which is NH3. So if you take NH4 plus, would you expect it to be acidic, basic, or neutral? Acidic. All right? At the equivalence point now, the pH has to be acidic. All right? And then if you write the overall reaction or the general form of the reaction, it will be where you have HCl, which is the strong acid reacting with ammonia, which is the weak base giving you NH4 plus. In fact, I'm writing the general form of the reaction. So it just turned out to be giving you salt here. All right? So if you write it in the general form, you see that what you're forming is ammonium chloride. You're forming the salt, the strong acid reacting with the base. So can everybody see that the general form or the overall reaction or sometimes they call this the molecular equation, all right, where it identifies the molecules? See that these are covalent compounds, so you can use the word molecular, all right? And this is also called the molecular equation. Okay? So now let's look at what the pH titration curve looks like. So this is what the titration curve looks like. So over here, we're measuring the pH on this side. On the y-axis, we have the pH of the analyte. Over here, we have the volume of the strong acid that's added. And you can see now the shape of this is very different. Okay? To begin with, you just have a weak base, all right? Before we add any strong acid, we have only a weak base in the flat. So we'd expect the pH to be high, all right? It's a basic solution. Now as you add a strong acid and you're neutralizing the weak base and converting it to NH4 plus, which is a weak acid, you know that the pH is going to go down. Okay? So what we see is that the pH decreases. But in this region, this is the buffer region, all right? Once again, over here in this region here, it would act as a buffer. So that's why it flattens out. The pH doesn't change dramatically there. It just kind of flattens out in plateaus because now what we have is in that mix, in that solution, you'll have NH3, which is the weak base, and you're going to have NH4 plus, which is its conjugate acid, and you have mixtures of both in substantial amounts. And so in that region, it's going to act as a buffer, all right? So that's where this flattens out. And then as you add more and more acid, now your buffer is going to get completely swamped and everything is going to get converted to NH4 plus. So at the equivalence point, now you can see all I have is NH4 plus, Cl minus, and water. Cl minus is the conjugate base of a strong acid. We know that's neutral. So Cl minus is neutral. Water is neutral, all right? And therefore, the only other species you have in there is NH4 plus, and the NH4 plus is the conjugate acid of a weak base. It's going to be acidic. And therefore, the pH is less than 7, all right? So you can see that the pH is less than 7, okay? Beyond this point, now you're swamped with a strong acid. So beyond the equivalence point, the pH is going to determine by the extra hydrochloric acid that's there, which associates completely. And you can see beyond the equivalence point, now rapidly the solution gets very acidic, all right? Now if this is the equivalence point, so you can see that in this titration, the concentrations are different, okay? In this titration, it turns out that the equivalence point is at 50 milliliters, all right? So half the equivalent point would be 25. So if you take 25 milliliters, that would correspond to that. And so at 25 milliliters, we know that we have a buffer there where the concentrations of the weak acid are equal to each other. So the concentrations are equal to each other, you know, it's a buffer. And if the concentrations are equal to each other, the ratio of acid to base would be 1. Log 1 would be what? 0. And therefore, we know pH equals pKa, all right? So here, even though we have a base and solution, this gives me the point at which pH equals pKa, all right? It's not pH equals pKa, even though we have an acid down there, some students say, well, we have a base down there, so why are we talking about kA rather than kB? But remember, pH always equals pKa, according to Henderson-Husselbalck equation, where the concentrations of acid and base are the same, all right? So here, this would be pH equals pKa. And so you don't even have to do any calculations. I ask you, what is the pH of the half-equivalence point? You just look at this table and you look for the conjugate acid-base pair that we're interested in looking at, and that would be ammonium and ammonia. And you know, pKa is 9.25, so what should this pH correspond to? 9.25, all right? Because we know that at the half-equivalence point, the pH has to equal pKa. Is that clear to everybody? So now we've looked at all three types of titrations. I did not do any calculations here, but you should be able to do the same type of calculations that we did for weak acid-strong base or strong acid-strong base. And I'd like you to try that at home. In fact, I've assigned a problem from your textbook where you get to practice that. Okay? Yes, go ahead. Can you explain what happened to the equivalent point, meaning the volume? So her question was, what happens at the equivalent point? So you have to turn to the net ionic equation. So in this titration, what's happening is the strong acid neutralizes the weak base, all right? And so essentially what happens is, when this reaction takes place, you have NH4 plus and water being formed. Okay? So when you have stoichiometric amounts of both or equivalent amounts of both, where all of this reacts completely with this and all of this reacts completely with this, so that you have no excess reagent left over, all right? Both reactants react completely. And that happens only at the equivalent point because you have to have stoichiometric amounts. You have to have the right amount of each so that you don't have any limiting reagent. You don't have any excess reagent. They react in stoichiometric amounts. So that only happens at the equivalent point, okay? So at the equivalent point, when both of the reactants have reacted completely, all that you have left behind is just the product. You don't have any excess reagent left over. So because all of them are reacted, all that you have there is just NH4 plus, water, and Cl minus, all right? The spectator ion is Cl minus. Cl minus is the conjugate base of a strong acid, and we know that that's neutral. We know water is neutral, all right? And so now we turn to NH4 plus. So those are only three species in solution. If you turn to NH4 plus, you know that that's a weak acid. It's the conjugate acid of a weak base, and therefore it's a weak acid. If it's a weak acid, it's going to be acidic, all right? And therefore the pH should be less than 7. Got it? Okay, so now we've completed looking at acid-based titrations, and now we're actually ready to move on to the last topic that we're going to look at in terms of titrations. So by now, you guys are experts at solving any equilibrium problem. And we focused, this entire quarter, we focused on equisiculabria, what happens in water, all right? And that kind of has been the focus all along. Now we're going to look at the last type of equilibria, and this deals with solubility, all right? So we're going to look at solubility product. And as all of you know, solubility deals with when a substance dissolves. So we're looking at solubility product in equisolutions, and so the word solubility is related to soluble when something dissolves, all right? So let's take something, as all of you know, we know that many, many ionic compounds dissolve easily in water, and we call them soluble salts. So if I take soluble salts. So we'll start by looking at soluble salts. And all of you know, when we use the word salts, we're talking about ionic compounds, and we said if you take ionic compounds that are soluble, what happens the instant you dissolve them in water, what happens to soluble salts? They dissolve, and when they dissolve, what does that tell you? They dissociate completely, okay? So give me an example of a really good soluble salt in ACL. We all know table salt in ACL is a very good soluble salt. So I start with an ACL, and all of you know that if I take a beaker, and I'm going to add solid sodium chloride. So you take, all of you know that ionic compounds tend to be crystalline solids, and if you take common table salt, you can see that they have nice crystalline granules. And so if you take a spoonful of that and put it in a beaker water and stir it, what's going to happen? It dissolves completely. And when we say it dissolves completely, we know that it has dissociated completely to give you Na plus and Cl minus, okay? Now what I'm going to do is I'm going to add another spoonful and stir it up. It's going to dissolve. I'm going to take another spoonful and stir it up. It'll dissolve. I'm going to keep doing this until the point comes when that solution is completely saturated with sodium chloride. So if I keep putting more and more and more sodium chloride, what I'll end up with is a really saturated solution. And the point at which this gets to be saturated, we know it's saturated because it can't take anymore. If you add another spoonful of sodium chloride, it's not going to dissolve anymore because that solution has taken as much as it can take, and therefore what will happen is you're going to have undissolved sodium chloride at the bottom. So we're talking about a saturated solution, a solution that can take as much sodium chloride as it can take, and now it's beyond the point where you can't dissolve anymore sodium chloride. At that point, you'll have solid material at the bottom, all right? Now when you have solid material at the bottom, what happens is that you will establish an equilibrium where you have Na plus equals plus Cl. In other words, you will establish an equilibrium where when one unit of NaCl dissolves, so one cation and anion dissolves, somewhere else they will deposit. One unit will deposit. You see that? So if the solution is saturated, then what happens is you will establish this equilibrium where you have the solid material dissolving, and you'll have, if the solid material dissolves at any point, you're going to have two dissociated cations and anions associating with each other and depositing on top of that solid, okay? And sometimes the solid material at the bottom is called, we call this a precipitate. So when you have undissolved material at the bottom, we say we have a precipitate at the bottom, okay? Now since this is an equilibrium, we should be able to apply the law of mass action to this scenario, but it turns out we can't because this solution is non-ideal. This solution is non-ideal, all right? And the reason it's non-ideal, the solution is non-ideal because it is not homogeneous. And the reason it's not homogeneous is that when the solution is saturated, the saturated solution has mixtures of ion pairs and ion clusters, all right? So the reason that the solution is non-ideal is because it's so saturated, now the cations and anions are so crowded by chance that they're so close to each other that they pair up. So we call that an ion pair. So when the solution is so crowded with cations and anions, now in solution you'll have one cation and anion so close to each other that they pair up and you have an ion pair. Or by pure chance if four of them are together, two pairs of cations and anions will form a cluster. You can have three pairs of cations and anions forming a three-pair cluster. You can have four pairs forming a four-pair cluster. So what happens is now you have all these different species in there. You just don't have cations and anions separated from each other. On top of that you have a pair paired up called an ion pair. You can have a few cations and anions forming clusters and the number of cations and anions in the cluster can vary. So the net result is now you have mixtures of different kinds of pairs and so we say the solution is non-ideal. Your solution has to be uniform. Your solution, if you're going to calculate concentrations and you want to determine how much of each species is there, if you have mixtures of all of this, you can have independent separate cations and anions. You can have in that solution not only that, but you can have pairs of ions. You can have one pair or you can have two, three, four, five pairs of ions clustered together and as a result the solution is no longer homogeneous. And because it's not homogeneous, we say it's a non-ideal solution. It's not an ideal solution is where everything is homogeneous, dissolved completely and all the species are the same. So ideal solution would be just having Na plus and Cl minus and that's it. But if you have clusters and ion pairs and so on, now it becomes non-ideal. So the net result is that the law of mass action cannot be applied. So when you take a salt that dissolves easily in water, a soluble salt, you can't solve an equilibria problem. You can't use the law of mass action because that solution is not ideal. But if you take insoluble, so we're going to call these insoluble salts, or we call them marginally soluble salts. So incompletely soluble salts, marginally soluble salts. There's another word that we use and I want and slightly soluble salts. So if you say insoluble salts, marginally soluble salts or slightly soluble salts. Now if you take these, they all mean the same thing. So sometimes we use the word insoluble, sometimes we use the word marginally soluble, sometimes we use the word slightly soluble. So these are the exact reverse soluble salts. All the salts are the ones that dissolve completely. Everything else falls under this category. And an example of this would be something like silver chloride. So silver chloride is a precipitate. It's insoluble and the reason it's insoluble is if you place it in water, it just sinks to the bottom. It just forms a precipitate. So now we don't need to form a saturated solution. The instant you put silver chloride down here, it will form a solid. It will form a precipitate. And because it's insoluble or marginally soluble or slightly soluble, a tiny, tiny, tiny amount actually dissociates. So it's like, this is the equivalent of like weak acids. Remember weak acids, weak bases dissociate only a small amount. The majority of it stays undissociated. The same thing here. So when you take an insoluble, marginally soluble, a partially soluble salt, now you set up this equilibrium. Now this is an ideal solution. Why? It's not saturated. This is a solution where only a very, very small amount of the silver and chloride dissolve. So in solution, they're far apart from each other. And so there's no chance that you can form ion pairs or ion clusters, okay? Because the amounts are so small and they're separated by billions and billions of water molecules that they don't form ion pairs or ion clusters. And so what you do is you establish this equilibrium where some solid will dissolve and then some of the dissolved material will combine to give you a solid and you would establish an equilibrium. And because this is an ideal solution now, we can apply the law of mass action. So since this is an ideal solution, the law of mass action can be applied. All right? So now only in the case of insoluble or marginally soluble or slightly soluble substances, we can apply the law of mass action. And if we can apply the law of mass action, then we can write the equilibrium we're going to look at, which is the solid precipitate in equilibrium with dissociated silver ions and chloride ions. Okay? Now can everybody see the difference between soluble salts and insoluble salts are their attraction to water. Soluble salts, the instant you put them in water, they dissociate to give you cations and anions because their attraction for water is greater than their attraction for each other. Insoluble, a margin of soluble salts, their attraction for each other is much greater than their attraction for water and that's why they form a precipitate. They don't dissociate in water, they'd rather stay associated as the solid. Okay? So if you look at silver chloride now, we can write, this is an equilibrium, but can everybody see that this equilibrium is different from the other equilibrium we looked at because we have a solid and we have the others dissolved in aqueous solution. When you have two different phases, what do we call that? Heterogeneous. And do you guys remember last quarter when we looked at the basic fundamental concepts of equilibria, when you deal with heterogeneous systems that have pure solids, pure liquids, what do we do? We don't include that in the law of mass action of the equilibrium expression. So in this case, if I were to write the equilibrium expression, which is the law of mass action, we'd say K and this has a special name, so we call the Ksp equals the silver ion concentration times the chloride ion concentration and this has a special name, this is called the solubility product constant. And remember the word solubility is the amount that dissolves, alright? And so you can see product means time, so what we're looking at is the solubility of silver and chloride, the amounts of silver and chloride that dissolve and you're multiplying that and that's what gives you the solubility product. See this is the solubility of AG plus, this is the solubility of chloride, this is the product of that and therefore we call it the solubility product constant. Because these are all substances that don't dissolve easily in water, would you expect Ksp to be a big number or a small number? Small. Small, because the equilibrium lies entirely on the solid side, only very, very minuscule amount actually will dissociate. So all Ksp values are really, really small. Remember the Ksp value is large, it'd be like sodium chloride, alright? If Ksp is really large, but then we can't apply Ksp to soluble substances because the solutions are non-ideal, alright? So we don't apply Ksp to soluble salts, we only apply them to insoluble, marginally soluble or slightly soluble, okay? So if you look at the value for this, this is about 1.6 times 10 to the negative 10. If you were to write the unit for this, you can see this is moles per liter, this will be moles per liter, so it will be moles squared per liter squared, okay? Often we, even though we know that the equilibrium constant has a unit, very often we omit that unit, okay? So you should remember that if it had a unit, this would be moles squared per liter squared, okay? And this would be at room temperature, alright? Now if I were to raise, remember we said that the equilibrium constant always depends on temperature. If you change the temperature, the value of this would change. And so if I raise the temperature to 100 degrees Celsius, now this value would be 2.2 times 10 to the negative 8. Now look at that and tell me, now does Ksp, as you raise the temperature, does Ksp increase or decrease? Increases. Remember, what does Ksp tell you? It's solubility product. It tells you what the solubility is. Whereas you raise the temperature, what happens to solubility? It increases. That means more dissolves. And we know that. If you want to dissolve a substance, you want to increase the solubility, what do you do? You heat it up, alright? And you can see when you heat up and increase the temperature, Ksp actually goes up. And if Ksp goes up, it tells us that the solubility has increased, okay? Now if you wanted to know, so let's take some other examples. So if I take another example, would be PbI2. This is an insoluble substance. The equilibrium you would set up would be Pb2 plus equus plus 2i minus equus. There are two i-bins there. So if you look at this formula, when it dissociates, it's going to give you 2i minus. However, in all of these equations, you have to maintain charge balance. So you can see we start with something that's electrically neutral. Therefore, this side has to be electrically neutral as well. Lead is 2 plus, and therefore it's got to have 2i minus as to balance that. And overall, 2 plus, 2 minus, they balance each other. And if you take the Ksp for this, this is of the order of 7.1 times 10 to the negative 9. If you take something like, and so if I want to write the law of mass action for this, this would be the concentration of Pb2 plus times the concentration of i taken to the power 2. If you take something like aluminum hydroxide, this is also an insoluble substance, the equilibrium you would establish is now, the way to figure out charge balance is to look at hydroxide. What is the charge on hydroxide minus 1? How many hydroxide are there? Three. So that means you have three minuses. To counterbalance that charge, you have only one aluminum. So what should the charge of aluminum come out to be? Three plus. So you have to figure out what the charge is. In every case, you have to figure out what the charge of your cation is and what the charge of the anion is. Now usually group one metals have a charge of what? Plus one. What is the charge of group two metals? Plus two. The transition metals are what? Variable. Variable. It changes. And so for transition metals, you don't know what it is. And if you want to figure it out, you always turn to the anion. So if you know the charge of the anion, you can figure out what the charge of the cation is. So in this case, aluminum comes out to be three plus and hydroxide. You need three hydroxides to balance that charge. And if you look at Ksp, Ksp comes out to be two times ten to the negative thirty-two. That's a tremendously small number. So you can see only a very, very tiny amount of that actually is soluble. The solubility is really, really small. And so if you want to write Ksp for that, Ksp would be the concentration of the aluminum cations times the concentration of the hydroxide ions taken to the power three. So because Ksp deals with substances that are insoluble, we know that all Ksp values have to be really small. You can never have a Ksp number that is greater than one, all right? Because that means there's a soluble salt and you can't have a Ksp for soluble salts, okay? So here is, I pulled this out of a handbook and it's really, really small and that's okay because what I really wanted you to focus on is that just like we have tables for Ka values, there are tables for Ksp values. And usually these are in alphabetical order, all right? And so you can see aluminum hydroxide and you can see the Ksp that corresponds to that. But what I wanted you to take a look at is, take a look at all the Ksp values. Can you see that all of these are numbers that are less than one because they're all insoluble, marginally soluble, or slightly soluble and they're all going to have really, really small numbers. Got it? So that's important to remember. Another thing that I want you to remember is that one way for you to identify whether a substance is insoluble or soluble is if it has a Ksp value that's really, really small, if it has a Ksp value at all, it has to be insoluble, marginally soluble, or slightly soluble. Another way to figure out whether a substance assault is soluble or not is to look at the solubility rules. In high school you have seen this before and we call these the solubility rules and this is a guide for you to figure out whether an ionic compound or salt will dissolve in water or not. So another way for you, rather than looking at Ksp values, another way to figure out whether an ionic compound or salt is soluble or insoluble, marginally soluble or slightly soluble is to follow these rules. Now I don't want you to memorize these rules but I do want you to be able to understand these rules and if you have the rules right next to you be able to apply them. So on an exam if I want you to use this I will actually provide the rules. I'm more interested in whether you can actually apply the rules rather than whether you are good at memorizing the rules. So the first rule is that all nitrates and acetates are soluble. So if you take an ionic compound where the anion is a nitrate or the anion is an acetate, a sodium acetate, potassium acetate, all of these will dissolve completely. So the first rule is that nitrates, ionic compounds with anions that are nitrates or acetates are soluble in water. Sometimes we write the formula for acetate like that, sometimes we write it as CH3CO2- they're the same thing. So CH3CO2- is what? The conjugate base of the weak acid is acidic acid. So acetate is the conjugate base of the weak acid acidic acid. The second rule deals with chlorides, bromides and iodides and what we know is all chlorides are soluble except when the cation is silver, when the cation is mercury and when the cation is lead. So except for those all chlorides are soluble. The same rule applies for bromides and the same rule applies for iodides. The third rule is all sulfates are soluble. So if your anion is a sulfate, all sulfates are soluble except barium, strontium and lead. They're all insoluble. Then they say calcium, mercury and silver are only slightly soluble. Remember for all intents and purposes we're going to say that insoluble mean the same thing as slightly soluble mean the same thing as marginally soluble. It's like categorizing them all as weak acids. Either they're weak acids or strong acids. In this case they're either they dissolve completely which are the soluble salts or the opposite. Yes, go ahead. What's the S after the ZA? That's a typo. Thanks for pointing that out. It's calcium sulfate. That's sulfur. There are two S's there and can you cross out the lower case S? So calcium sulfate, mercury sulfate, mercury sulfate and silver sulfate are only slightly soluble. The third one is all group 1 metal ions and ammonium ions are soluble. So if you have sodium potassium or if your cation is ammonium they're soluble in water. All hydroxides are insoluble except group 1 metal ions and ammonium ions. Barium, strontium, calcium, these are group 2 metals. Barium, strontium, calcium are marginally soluble. The set of rules are on the class website so you can print it out. So I put it in, if you look at yesterday's date I put this up already so you can print it out. Six is all sulfides are soluble except group 1 metal ions and ammonium. Calcium ions, strontium ions and barium ions are also insoluble. All phosphates and carbonates are insoluble except group 1 metal ions and ammonium ions. Most chromates are only slightly soluble except those are group 1 metal ions and ammonium ions. Okay? So this covers almost every anion or cation combination that you can come across. All right? Now what I want to keep in mind is there are some very rare instances where you may have two rules that give you kind of, it's very, very rare, that two rules will give you opposite information. All right? If that does happen then what you have to make keep in mind is any rule on top has precedence over any rule at the bottom. Is that clear? It kind of goes in that order and if you encounter it's very, very rare. It's very unlikely you're going to encounter it but if you ever encounter where two rules kind of give you conflicting information then always make sure that anything on top takes precedence over anything below. Got it? So either you can use KSP values or you can use solubility rules to figure out whether anionic salt is soluble or not. Okay? So now we can, now that we understand the difference between soluble salts and insoluble salts and that why we focus only on the insoluble or the marginally soluble or the slightly soluble compounds now we're ready to look at solubility and KSP. All right? So we want to look at the relationship between solubility and KSP. We said before if the KSP gets bigger or any substance that has a KSP that has a bigger number we know that's related to solubility so that means that substance will be dissolving a little bit more than something that has a lower KSP. Okay? So solubility refers to the amount of the substance that dissolves. Okay? So if we want to look at solubility and let's take an example of silver chloride. So we know silver chloride dissociates to give you silver, cations and chloride anions and we know this equilibrium is given by KSP and the equilibrium constant is a really, really small number. Now we want to calculate what is the solubility? How much of this actually dissolves? All right? To do that we don't need to go through the complete ice chart because you look at, you usually use the ice chart where you look at the initial, then the change and then at equilibrium if you need the initial concentration. Now if you look at this, am I interested in the initial concentration of silver chloride? It doesn't dissolve in water anyway. All right? It doesn't appear in the law of mass action. So all I'm interested in looking at is what do I have at equilibrium? Can everybody see at equilibrium if I look at the concentrations? We know that if negative x dissociates we know plus x plus x would form. So at equilibrium all I'm left with is just that, xx. Can everybody see that? So all I'm interested is the amount of silver and chloride at equilibrium and since they dissociate in a one to one stoichiometry that's going to be x. So I know that Ksp equals 1.6 times 10 to the negative 10 which gives me the silver ion concentration times the chloride ion concentration which is x times x which is x squared. Okay? So if I take the square root of that I end up with 1.3 times 10 to the negative 5 moles per liter. Now this is the amount that dissolved. Remember this is the amount that's consumed. This is negative x and the amount that's so. Since this is in moles per liter we call this molar solubility. All right? So we have in general solubility and this gives the amount or concentration of AgCl that dissolved. All right? Remember solubility is related to how much actually dissolves, how much is soluble and therefore when you talk about molar solubility it gives us in terms of concentration the amount of the starting material that dissolved which is x. Now sometimes it's customary to give this as gram solubility. So if you want to give this in terms of gram solubility so here this is molar solubility and if it's molar solubility the unit is moles per liter. All right? If you give it in grams solubility you want to know how many grams of the solid material actually dissolved. All right? And so if I want to convert moles to grams what do I need to do? I need to multiply by the molar mass. All right? So I'm going to start with 1.3 times the n to the negative 5 moles per liter and I'm going to multiply by the molar mass of silver chloride. So you go look at the molar mass of silver, the molar mass of chloride, add them together that gives me 143.3 grams of silver chloride per mole. So this is the molar mass, this is times the molar mass of AgCl which is the reactant because solubility refers to the amount of the reactant that dissolved. Okay? So you can see moles and moles will cancel out and so I end up with 1.9 times 10 to the negative 3 grams per mole of AgCl. So you can calculate solubility in terms of concentration and that is x, that's the amount that dissolved, that would be molar solubility or you can calculate in terms of grams and that would be gram solubility. Oh I'm sorry grams per liter, yes. Alright so the unit would be grams per liter, grams per mole would be the molar mass. Got it? So let's apply this in a problem. So let's take a very straightforward and easy problem and so if we take this example where the solubility product constant for calcium fluoride is given to you. Okay? So you calculate the concentrations of calcium and fluoride in a saturated solution of calcium fluoride at 25 degrees Celsius and determine the solubility of calcium fluoride in grams per liter. Now here they're using the word saturated and when you talk about insoluble substance the saturated refers to just the teeny tiny amount that will dissolve. Alright? When you talk about soluble salts if you use the word saturated it means that it's super super saturated. Do you get it? So that's the difference. So you should be able to recognize from the KSP that's 10 to the negative 11 and so this is an insoluble substance so very little actually dissolves and when the maximum amount that can dissolve is dissolved we call that a saturated solution but the maximum amount that an insoluble salt dissolves is going to be really a minuscule amount. Got it? So we're looking at calcium fluoride so we're going to set up the equilibrium so we know this is insoluble and we're looking at KSP so the equilibrium that you're going to establish would be this. Calcium is group 2 so it's got to have 2 plus, fluoride is group 7, it's got to have minus 1 charge, there are 2 of those and we know KSP equals 3.9 times 10 to the negative 11 that number is going to be really really small. That means that if I want to calculate, I know KSP, if I want to calculate the concentration of calcium and fluoride I know at equilibrium if X is consumed how much calcium would form X but if X is consumed how much fluoride would form to X. Okay so don't forget that and therefore we know that KSP equals calcium ion concentration times fluoride ion concentration raised to the power 2 which is X times 2X squared. Make sure you put this within parenthesis because if you take 2X squared that becomes 4X squared times X, what would that give me? 4X cubed. So we know that 4X cubed should equal 3.9 times 10 to the negative 11 which is KSP therefore if you work out the value of X it comes out to be 2.1 times 10 to the negative 4 moles per liter. So X is solubility, this is molar solubility, this is the amount of solid that dissolved. The first part is they want us to calculate the calcium ion concentration so therefore the calcium ion concentration is actually X and that would be 2.1 times 10 to the negative 4 molar. If you take the fluoride ion concentration the fluoride ion concentration would be what? 2X and therefore it should be 4.2 times 10 to the negative 4 molar. Now lastly they want you to calculate the solubility of calcium fluoride in grams per liter. So now we want to calculate gram solubility and that would be the molar solubility which is 2.1 times 10 to the negative 4 moles per liter times the molar mass of calcium fluoride. So look at the periodic chart, calculate the molar mass of calcium plus 2 fluorides and if you add them up the number comes out to be 78.1 grams per mole, moles and moles are cancelled out and this is calcium fluoride. And therefore what you end up with is 0.016 grams per liter of calcium fluoride. So the gram solubility gives you the mass of calcium fluoride that actually dissolved per liter and that we figured out. So next class we're going to take a few more examples of solving equilibrium problems associated with KSP. So we'll stop there for today.
to look at three types of acid-base titrations. look at the third scenario where you have a weak we focused, this entire quarter, we focused on equilibria, and this deals with solubility, all concentration and this has a special name, this solubility product. See this is the solubility of