[NOISE] All right, good afternoon. [NOISE] Today is November 14, 2019, which means exam number 3 is now just seven days away. Everybody know what to expect? Everybody looked at the practice exam. Come on, come on. Got to do these things. What I want to do today is pick up on where we left off last time talking about acid-base chemistry. Last time we were showing how you can use equilibrium techniques to figure out pH of acidic or basic solutions as long as you do the equilibrium constants K_a for a weak acid, K_b for a weak base. What I want to talk about today is the concept called conjugate acids and conjugate bases, and what we mean by the conjugate of an acid or a base is what's left after the acid donates its proton or the base, accept its proton. Let me show you this chart, and there is a similar chart that appears in your textbook. What they done on this chart is list acids in the left-hand column, bases in the right-hand column. As we indicated before, red means acid, blue means base. But they're also trying to show you two things, but it's chart. One is acids and bases that are conjugates of each other. In other words, if you think about HCl behaving as an acid, okay, what do acids do? They donate protons. After HCl donates its proton, what's left? CL minus, of course. After HF donates its proton, what's left? F minus after NH4 plus donates a proton, what left NH3 and so on. The point is everything in this column as an acid in the sense that they can donate a proton. What's right next to it in the base column is what's left after it donates its proton. Each of these is called the conjugate base of the acid that appears in this column. Conversely, if you want to think of these as bases, what's formed after they accept the proton, the conjugate acid. The point is, a conjugate acid or a conjugate base differs by one proton from the original acid or base. The conjugate base has one less proton than the original acid does, the conjugate acid has one more proton than the original base does. Now, suppose HA represents some weak acid. HA reacts with water to form hydronium ions and leave behind A minus after HA donates its proton. The point is HA behaving as an acid generates A minus as its conjugate base. Water in this reaction is behaving as a base in that it's accepting a proton. So the conjugate acid of water is the hydronium ion H_3O plus. One way of thinking about this, since this equilibrium, think about which one would be the acid and which one would be the base for the reverse reaction. In this case, this would be the acid, this would be the base. But I've used the red and blue highlighting here to say that A minus is the conjugate base of the acid HA, H_3O plus is the conjugate acid of the base, water. When we say that A minus is the conjugate base of the weak acid HA, we don't necessarily mean that this is a strong base, but we do mean that it is strong enough to do this with water. A minus this can react with water to generate hydroxide ions and form some HA again. So in this reaction, water is behaving as an acid, A minus behaving as a base. The conjugate acid of A minus is HA, the conjugate base of H_2O is the hydroxide ion. Both of these are equilibrium. Now, let me just go back to this table for a moment and let you get back to work copying there. The other thing they're trying to show you on this table with the color-coding. Up here at the top of the asset column, we have things like HCl, sulfuric acid, nitric acid, things we identified last time as being strong acids. Dark red for strong, as we go down the column, we get paler red, pink. Finally, not much noticeable red. The point they're trying to make is as she go down the column here, the acids get weaker. But no surprise, that means their conjugate bases become stronger. We go from nothing to pale blue, to darker blue, as we go down the column here. If you think about it, this should make sense, when we say that HCl is a strong acid. That means it has a strong tendency to get rid of its proton. What does that say about the tendency of the chloride ion to hang on under that proton. Relatively weak, right? By contrast, a weaker acid like HF, well, if HF is a weak acid, that means that the fluoride ion has a somewhat stronger tendency to hang onto a proton, then a chloride ion does. You can actually quantify this statement. If you were to write equilibrium expressions for each of these two equilibria that we were just showing you a moment ago. In this inner parentheses, we have the equilibrium expression for the first equilibrium hydronium ion concentration times A minus divided by HA concentration. In the second set of parentheses, we have the equilibrium expression for the second equilibrium, concentration of HA times concentration of hydroxide ion divided by concentration of A minus. Notice that in both cases we ignore water, which is a pure liquid. The point is this represents K_a for the acid, and this represents K_b for the conjugate base of that acid. When you multiply these two together, some interesting things happened. Concentration of A minus cancels out the numerator and denominator. Concentration of HA cancels out numerator and denominator. This simplifies to the concentration of hydronium ion times the concentration of hydroxide ion. But we saw last time that for any aqueous solution, the concentration of hydronium times the concentration of hydroxide is going to be equal to the water constant 1.000 times 10 to the negative 14. But this expression is simply K_a for our weak acid, HA and this expression is K_b for the weak base, that is its conjugate, K_b. While it is true that the stronger an acid is, the weaker its conjugate base is, but I encourage you to use the terms stronger and weaker as opposed to strong and weak. You'll see why in the example we're going to work through in a few moments. Likewise, the stronger a base is the weaker its conjugate acid would be. If you want to think about this in more mathematical terms, as K_a becomes larger, which means the acid is getting stronger, its conjugate base K_b must be getting weaker in order for the product of the two to come out to be equal to 1.00 times 10^negative 14, which must be true for any acid and base that are conjugates of each other. Do these concepts make sense? I'll let everybody finish with this slide. Work out some other details on some sample problems that we'll show you. In fact, the sample probably we have coming up is on page 145 in the lecture notes for those of you who are following along. [BACKGROUND] [NOISE] Would anybody like more time with this slide? Let's apply this to the problem that appears on page 145. Chemists sometimes use the term salt of a weak acid or a weak base to refer to the conjugate of that weak acid or a weak base. The title of page 145 in electron notes is calculating pH for solutions of salts of weak acids. By that we mean conjugates. Here's a problem that appears on that page. Last time we saw that acetic acid, which has the formula HC_2H_3O_2, has a K_a value of 1.8 times 10^negative 5. What would be the pH of a one molar aqueous solution of, now here's where you have to read the question carefully. Not acetic acid but sodium acetate. Now, before we try any mathematical solutions to this problem, let's think qualitatively for a moment. Large number or a small number here? Small number. Small number. Therefore, is acetic acid a strong acid or a weak acid? [NOISE] Small value of K_a tells you that it is a? Answer's staring you in the face on the screen Weak acid. Weak acid, yeah. Small value of K_a means weak acid. If this were a large number like 1.8 times 10^5, then it would be a strong acid. Given that this is a weak acid, what does that tell you about its conjugate base? Relatively strong. Strong enough to react with water as shown here. Now, first of all, when you dissolve sodium acetate in water, you get sodium ions and acetate ions. The sodium ions don't do very much. They float around in solution. We talked about spectator ions earlier in the semester. Sodium ions are spectator ions here, but the acetate ions do have an impact on the pH of the solution, because since they are the conjugate base of a weak acid, they are strong enough to react with water at least to some extent and generate acetic acid and hydroxide ion. The point is, we are generating hydroxide ion. Is this solution going to be acidic or basic? Basic. Basic. The Arrhenius definition of a basic solution or a base, is any substance which when dissolved in water generates hydroxide ions. We're generating hydroxide ions and making a basic solution, the pH is going to be what? Seven, greater than seven or less than seven? Greater than seven. Greater than seven for any basic solution. Well, as you might have surmised, this being an equilibrium problem we're going to use an ice table to get help solve it. As usual, the initial column means that we have dissolved the sodium acetate in water. We have acetate ions floating around in the water, but we assume that none of this has taken place yet. What's the initial concentration of the acetate ion? [NOISE] Bless you, but that's not the right answer. [LAUGHTER] Yes. One molar. One molar. As we said up here, we have one molar sodium acetate. That means when it dissolves in water, we get one molar sodium ions, one molar acetate ion. One molar acetate ions to begin with. What's the initial concentration of the other two? We're going with to assume zero with the usual asterisk next to the hydroxide because we actually know that in a glass of water we actually get 1 times 10^minus 7 molar hydroxide ion. It's not quite zero, but we're going to assume that that's pretty much negligible compared to what we get from the reaction taking place. What's the change column going to look like? Take a look at the balanced equation. Tell me what the change column is going to look like. Yes. Minus x plus x plus x. Minus x plus x plus x. These two concentrations have to go up. They can't go down, they're starting from zero. They're both going to go up by the same amount because it's a one-to-one ratio. Where do those things come from? This thing, grabbing protons away from water molecules. This concentration goes down by x, the other two go up by x. The equilibrium column is just the sum of the first two columns. 1 minus x for acetate, x for acetic acid, x for hydroxide ion. Now the one thing we haven't done yet is write the equilibrium expression for this equilibrium. What's the equilibrium expression for this equilibrium going to look like? [NOISE] Yes. X times x divided by 1 minus x. Well, but I was thinking in terms of these concentrations over here. You're right, but let's back up and do the concentration terms first. HC_2H_3O_2 times OH divide by C_2H_3O_2 Yes. These two are going to be in the numerator. This one's going to be in the denominator. As usual, we ignore water because it's a pure liquid. Any pure liquid or a pure solid does not show up in the equilibrium expression. Before we show you that, we would like more time with this slide. Everybody good? [NOISE] The point is, the equilibrium expression is going to look like this, and we'll put the term from the ICE table into this in just a moment. This is going to be equal to K_b. We know that it's a K_b expression because we're generating hydroxide. You told me a moment ago that the pH of this solution should be greater than seven, should be a basic solution. However, we don't have K_b for the acetate ion, what we have instead is K_a for acetic acid. How do we find K_b? Say that louder. K_a times K_b is equal to one times 10^ negative 14. If you're going to sit in the front row, you know I'm going to be reading your lips, and they are going to say it. Yes, the point is that for any acid and base that are conjugates of each other, K_a for the acid times K_b for the base must be equal to one times 10^ minus 14. To find K_b for the acetate ion, we take one times 10^ minus 14, and divide that by K_a for acetic acid, which is 1.8 times 10^ negative five. This comes out 5.6 times 10^ negative 10. What we want to do now is take the information from the ICE table, and plug it in, and solve for x, which turns out to be extraordinarily easy in this particular case, because concentration of acetic acid is x, concentration of hydroxide ion is x. Concentration of acetate ion is one minus x. All of that is equal to K_b. Is this a large number or a small number? Small. Very small number. We could considerably rearrange this into a quadratic equation, and solve it on that basis. However, this is a very small number. What that tells us is that x is going to be a very small number also. So much so that subtracting X from one is really not going to make very much difference, so we're going to ignore X in the denominator. Which means that the denominator becomes one, which means that this whole fraction just simplifies down to X squared. Now, the worst thing we have to do to solve for X is take the square root. Square root of 5.6 times 10^ negative 10, it's 2.4 times 10^ negative 5. Whenever you make an assumption to try to solve a problem, it's always a good idea to check to see if your assumption is valid. The assumption was that one minus X would be approximately one. If we subtract 2.4 times 10^ negative five from one, we're going to get 0.99996, something like that. Is that pretty close to one? Yeah, so our assumption is valid. What does x represent from the ICE table? [NOISE]. Change. Somebody said? Change. Well, it does represent something for the change column, but I was thinking more in terms of looking at the equilibrium column, the point is we're trying to figure out what the pH of the solution is. What does x represent that's going to help us figure out what the pH of the solution is. Concentration of hydroxide. Concentration of hydroxide. That's the important point. We know what the concentration of hydroxide is. By the way, our other assumption that the concentration of hydroxide that comes from the auto photolysis of water, one times 10^ negative 7, pretty much negligible compared to this. If we add one times 10^ negative 7 to this, we had 2.41 times 10^ negative 5. Again, not much difference. When you know the concentration of hydroxide, how do you find the pH? [NOISE]. Negative log. Take the negative log, and then what? Subtract from 14. Subtract it from 14. If you take the negative log of this number, you have not found pH, you have found pOH, which in this case turns out to be 4.62. But finding pH is straightforward from there, because the sum of pH, and pOH must be 14. Subtracting 4.62 from 14 gives us 9.38. We predicted earlier that this would be a basic solution with a pH greater than seven, and that's in fact what we're seeing here. [NOISE] Are there any questions about any aspect to this problem? Yes. What is the threshold for [inaudible]? The threshold for this assumption here? Yes. I would say approximately one times 10^ negative 4. If it's smaller than that, you can probably go ahead, and make the assumption. If it's greater than that, you may be stuck with solving a quadratic equation or something like that. That usually works. I could send you home to figure that out yourself. Solve the problem both ways, and then see if you get the same answer doing it by both techniques, but you asked, so that's my answer. Any other questions about any aspect of this problem? If all of this makes sense to you then I think you probably have a reasonably good grasp. Some of the most important number crunching calculation techniques that are helpful for solving acid-base problems, such as those that you will find in Chapter 14 in your textbook. Would anybody like more time with this slide? Section 14.4 in your textbook is titled hydrolysis of salt solutions. That's actually what we were just talking about. [NOISE] Again, reusing salt here in the context of the conjugate of a weak acid or a weak base as opposed to just sodium chloride. But the term hydrolysis, you may have heard of before. Hydro means water. Lysis means cleaving or tearing apart. What's meant by hydrolysis in this context is the fact that when you dissolve some ionic compounds in water, you get solutions with a pH of seven. But when you dissolve other ionic compounds in water, you get solutions with a pH of something other than seven. Maybe acidic, maybe basic, depending on what the salt in question happens to be. What's meant in this context by the hydrolysis of salts is a reaction taking place between some ion that results when you dissolve the ionic compound in water, and the water itself to generate either hydronium ions or hydroxide ions. If you're generating hydronium ions, you're making the solution acidic, and if you're generating hydroxide ion, you're making the solution basic. If we speculate that HA is a weak acid, then its conjugate base, A-minus, is going to be strong enough as we just saw in the last problem, to react with water to some extent, and generate some hydroxide ions, which is going to make that a basic solution. By contrast, if B is a weak base, then its conjugate acid, HB plus, is going to be acidic enough to react with water to some extent, and form hydronium ions, therefore giving rise to acidic solution. One thing we're going to look at is how you can look at the formula of a compound, and figure out by just looking at the formula whether dissolving this compound in water is going to give you an acidic solution, a basic solution, or a neutral solution. Some of this does have to do with having an understanding of which acids are strong acids? Which bases are strong bases? Therefore, by default, which acids are weak acids? Which bases are weak bases? Will show you some examples in a moment. There's a technique that I think you might find helpful. Well, we'll just go ahead, and show you the examples in a moment, and you'll see what's going on. [BACKGROUND] [NOISE]. That's HA plus OH minus. This part right here? Yeah. Yeah. HA plus hydroxide has a negative charge with a circle around it. [BACKGROUND] Does everybody have what they need from this one? [NOISE] The examples we're going to show you here, the idea is to look at the formula of the compound and try to figure out whether it will form an acidic solution, a basic solution, or a neutral solution when you dissolve it in water. To figure that out, here's one question you can ask yourself. Which is stronger, the parent acid or the parent base? By that I mean, that in principle, any ionic compound can be formed by mixing together an acid and a base in the correct proportions. Think about what the original acid and the original base must have been and then try to figure out which one of those two is stronger. That might give you a clue as to whether the solution is going to be acidic, basic, or neutral. We'll start with a very simple example. Sodium chloride, saltwater. Acidic, basic, or neutral? [BACKGROUND] Somebody said, maybe not. Anybody want to guess? Yeah. Acidic. No. Somebody else said? Neutral. Somebody else said neutral and neutral is the right answer. Here's how you figure it out. What acid and what base do you have to mix together to make sodium chloride? What will be a good choice for the acid? HCl. HCl; the chlorine has to come from somewhere. What would be a good choice for the base? Sodium hydroxide. Sodium hydroxide. If that's not obvious, here's what you can do. Write down the formula, NaCl, split it in half, positive ion on one side, negative ion on the other side [NOISE], and then add water. But by adding water, I mean at H to this side, add OH to that side, HCl, NaOH. HCl, strong acid or weak acid? Strong acid. Strong acid. Sodium hydroxide, same question, it's a base. Strong base or weak base? [BACKGROUND] I'm going to tell you this one more time. Water-soluble hydroxides are strong bases. Sodium-anything is water-soluble, remember the solubility rules? HCl is a strong acid, sodium hydroxide as a water-soluble hydroxide is a strong base. In trying to address the question which one is stronger, in this case, the answer is neither one. Because by definition, a strong acid reacts 100 percent. A strong base reacts 100 percent. One hundred percent and 100 percent are equal to each other. If they are equally strong, it's neutral. Does the logic make sense? Let's try another example. Sodium acetate; NaC_2H_3O_2. Is the solution going to be acidic, basic, or neutral? Basic. Say again. Basic. Basic, did you say with a question mark at the end of it? The answer is basic. [NOISE] Which should have been obvious because we just did a problem or we figured out that the pH of a one molar sodium acetate solution was something greater than seven. But let's reason it out. What acid, what base would you mix together to make sodium acetate? What's a good choice for the acid? Acetic acid. Acetic acid. What's a good choice for the base? Sodium hydroxide. Sodium hydroxide. Again, if that's not obvious, write down the formula, cut in half, H to this side, OH to that side. We know that sodium hydroxide is a strong base. What about acetic acid? Strong acid or a weak acid? Weak acid. Weak acid. The reason the solution is basic is when you mix together a weak acid and a strong base in equal molar amounts, the base is stronger than the acid, therefore the solution as a whole turns out to be basic. You buying this? [BACKGROUND] Make sense? I've included here the equation for the reaction that we saw in the previous problem to explain why it's basic. The point is that the acetate ions, once you turn them loose in water, react with the water and generate hydroxide ions. If you're generating hydroxide ions it's going to be a basic solution. Ammonium chloride, NH_4,Cl. Acidic, basic, or neutral solution when you dissolve it in water? Acidic. Acidic is the right answer. How did you come up with acidic? [NOISE] I bet I know how you came up with acidic. I had already done one neutral, one basic, the next one's going to be acidic. Suppose you didn't see that, how would you have figured out that this was acidic? What's the acid, what's the base that you mix together to do this, way back there? [BACKGROUND] We have chlorine, so that's going to be HCl and then what do we mix together with the HCl? >> NH3 is the right answer. Now, I'll just point out one thing about this little splitting the formula thing that we are doing here. If you write this one down and cut the formula in half and do H and OH, like before, on this side you get ammonium hydroxide. Well, the problem is there is really no such thing as ammonium hydroxide. Bottles that are labeled ammonium hydroxide are really just aqueous solutions of ammonia. The point is, the original acid and base are in fact HCL and ammonia. HCL, of course, is a strong acid. Ammonia, we saw it's KB earlier, 1.8 times 10 negative 5, that's a small number, so it's a weak base. Mix together a strong acid and a weak base. The acid is far greater than the base, therefore the solution will be acidic. The reason why it's acidic is that the ammonium ions NH4 plus that are floating around in solution, react with the water molecules and generate hydronium ions, H3O plus. But the point is if you are generating hydronium ions, you are making an acidic solution. Are there any questions about any of these examples? While people are finishing up with this slide, we're going to show you a few more examples. On this slide, we discussed three possible scenarios. Strong acid, strong base, which gives us a neutral solution, weak acid, strong base, which gives us a basic solution, strong acid, weak base, which gives us an acidic solution. The fourth possibility, what happens if the original acid and the original base are both weak? What would you guess? Neutrals, not a bad guess, and that's true sometimes. But what makes it a little bit more complicated? The keyword here is not strong, it's stronger. Is everyone done with this slide? Let me show you what I mean by stronger being the key concept here. If the original acid and the original base are both weak, you cannot automatically assume that it's neutral. What you have to do is come up with some quantitative expression of how weak is weak. In other words, what you have to do is check KA for the acid and KB for the base. Let's consider ammonium acetate. Well, from the previous slide, we know that ammonium ions come from ammonia reacting with an acid, and acetate ions come from acetic acid reacting with a base. The original acid and the original base, are acetic acid and ammonia, both weak. Now, we got to do is look at the KA and KB values. KA for acetic acid, 1.8 times 10 to the negative 5, KB for ammonia, 1.8 times 10 to the negative 5. This turns out to be neutral, but the reason it turns out to be neutral is that these two numbers are the same. In other words, acetic acid is just about exactly as weaker acid, as ammonia is a weak base. The answer in this case is neither one is stronger because KA is equal to KB, and that's why it turns out to be a neutral solution. But let's look at the next example, which is ammonium fluoride. Pretty sure the original base is going to be ammonia, what's the original acid going to be? HF, hydrofluoric acid. If I tell you that KA for hydrofluoric acid is 3.5 times 10 to the negative 4. Here's KB for ammonia. Is a solution of ammonium fluoride going to be acidic, basic or neutral? Acidic, because anything times 10 to the negative 4 is larger than anything times 10 to the negative 5, KA is greater than KB. That leaves the original acid, although it's a weak acid is stronger than the original base. That's why the solution tends to be acidic. Does the reasoning makes sense? One more example. Ammonium cyanide, NH4CN. Ammonia is going to be the base, what's the acid going to be? Formula for the original acid, HCN. It's called hydrocyanic acid, but we wouldn't expect you to know that. But we do expect you to figure out, is that if we tell you that the KA for hydrocyanic acid is 4.9 times 10 to the negative 10. That is our ammonium cyanide solution going to be acidic, basic or neutral? How can you tell? Basic because the value of KA is less than the value of KB. Yep. In this case, the original base, although it's weak, is stronger than the original acid. That's why when you create this salt by mixing those two together in equal molar amounts and dissolve it in water, you get a basic solution. The moral of this story is that some ionic compounds give acidic solutions, and some give basic solutions, we dissolve them in water, some of them do in fact give neutral solutions. In principle, if you know something about what acids are strong, what bases are strong, what acids are weak, what bases are weak, KA and KB values, if necessary, you should be able to figure out whether the solution that you're about to make by dissolving the ionic compound in water is acidic, basic or neutral. Questions about any of these examples? >> Who needs more time with this slide? [BACKGROUND] Section 14.6 is titled simply Buffers. Who knows what a buffer solution is and what it does? [BACKGROUND] Have you ever heard this word before? [BACKGROUND] May know what a buffer solutions are all about? Well, then, [NOISE] the purpose of a buffer solution is to try to maintain a relatively constant pH. The way buffer solutions do that, is of what they are composed of, is some weak acid and its conjugate base. For example. Suppose we have some weak acid HA, which when you dissolve it in water, does this by definition. But you also have some of its conjugate base A minus dissolved in water, which means it does this by definition. The point is in a buffer solution you have both of these equilibria happening at the same time. Now, when I say that buffer solutions resist changes in pH, here's what I mean. Suppose you have both of these equilibrium going on in a buffer solution, and then you add some strong acid like HCl. Well, when you add HCl to water, it generates hydronium ions. Let's apply Le Chatelier's Principle. If we generate hydronium ions, which way is this equilibrium shift? [BACKGROUND] To the left. Those extra hydronium ions react with A minus to form a weaker acid HA, along with water. But the main point is what you've done in the process of doing that, is convert this strong acid hydronium ion into a weak acid, HA and that lessens the change in pH. We'll show you some examples in a moment. Suppose we had dumped in some strong base like sodium hydroxide. What does that do to this equilibrium? Shift to the right or shift to the left. Thank you. We add hydroxide on the right side, we shift to the left side. The added hydroxide reacts with the acid HA, degenerate the weaker conjugate base A minus along with water. In the process, you convert a strong base hydroxide into a weaker base, A minus, which minimizes the impact on the pH. Now, let me give you an idea of just how big an impact this can have. What's the pH of regular water, assuming it's distilled and purified and all that? [BACKGROUND] Seven. Suppose we have one liter of pH 7 water. Now normally when we see the formula HCl, we think, hydrochloric acid, and that's a solution. But HCl itself, hydrogen chloride is actually a gas. The way you make hydrochloric acid solution is to dissolve HCl gas in the water. Suppose we take one-tenth of a mole of HCl, and dissolve it in our one liter of water. Let's assume that the volume remains at one litre. Well then what we've done is create a 0.100 holder aqueous HCL solution. Since HCl is a strong acid, how would you find the pH of that solution? Yeah. It's going to end up. If not, that's okay. Yes. >> Take the negative log of the concentration. >> Take the negative log of that concentration because for a strong acid, the hydronium ion concentration is going to be the same as the concentration of the acid. pH, they get a log of hydronium ion concentration. In this case, negative log of this number, turns out to be one. Now, let me remind you of something from a diagram that we showed you previously. pH is a logarithmic scale. Every time you change by one pH unit in either direction, you change the acidity or basicity by a factor of 10. Adding 0.1 mole of HCl to our 1 liter of pH 7 water to create a pH of the solution, changes the pH by 6 pH units, which means this is 10 to the 6th power, or a million times more acidic than this is. Remember that as we look at the second part of this problem which talks about what happens when you have a buffer in there. They are just a [inaudible]. As it is people have done with this slide. Anyone need more time here? [NOISE] At the bottom of page 827 in your textbook, [NOISE] they refer to an equation named after the two people who first thought it would be a good idea. It's called the Henderson-Hasselbalch equation. Let me just take a moment to show you where this equation comes from, which they do in your textbook. But I'll walk you through the algebra. What the Henderson-Hasselbalch equation is mostly used for is to find the pH of buffer solutions, which by definition contain both a weak acid and its conjugate base. Here is the Ka expression for some weak acid HA. What I've done on the next line is to take the negative logarithm of both sides. Then to the right side, I'm going to apply one of the fundamental laws of logarithms, which says that the logarithm of the product of two numbers is the logarithm of one plus the logarithm of another. What that allows us to do is drag it out negative log of H_3O plus concentration by itself, and then subtract negative log of A minus over HA from that. [NOISE] Well, there's a reason for this, of course, what's the negative log of a hydronium ion concentration, better known as pH. Negative log of Ka is called pKa. The P in front means take the negative log of, so pKa means take the negative log of Ka, pH means take the negative log of H_3O plus. We could do it here too, but we're not. The last step in this to arrive at the Henderson-Hasselbalch equation as it's usually written is just add this second term to both sides to get pH by itself. [NOISE] The usual form of the Henderson-Hasselbalch equation is pH equals pKa plus the LOG logarithm of the ratio of the A minus concentration divided by the HA concentration, where HA is the acid and A minus is the conjugate base of that acid that make up whatever your buffer happens to be. The point is that since A buffer solution contains both a weak acid, HA and its conjugate base, A minus, if you know both of those concentrations, then you can figure it out fairly easily what the pH of the buffer solution is going to be as long as you also know Ka or pKa. If you know Ka, it's easy to find pKa. Just take the negative log of it for whatever your weak acid happens to be. We'll show you how this applies to the situation we were describing a few moments ago as soon as everybody is done here. [NOISE] Everybody okay here? [NOISE] Anybody would like more time with this slide? Remember the equation of the green box as we look at the next slide. How would you find the pH of a buffer solution created by dissolving one mole of sodium acetate and 0.1 mole of acetic acid in enough water to form one liter of solution. Well, first of all, this buffer is based on acetic acid. You would of course have to know the Ka for acetic acid. We saw that earlier today. Ka for acetic acid is 1.8 times 10^negative 5. How would you find pKa? What does that p in front mean? Take negative log. Take the negative log of this number. So pKa is simply the negative log of Ka, negative log of 1.8 times 10^negative 5, which turns out to be 4.74. The thing that drives most people the craziest about using the Henderson-Hasselbalch equation, keeping straight which one's which? I think the easiest thing to do is look at the formulas. Whichever one has the extra hydrogen, that's the acid. Whichever one doesn't is the conjugate base. In this case, acetic acid has that extra hydrogen, so acetic acid is our HA. Sodium acetate, you really don't have to pay attention to the sodium here because it's just going to be a spectator ion. It's the acetate ion that has an impact. The A minus is going to be the sodium acetate. Here's the Henderson-Hasselbalch equation. What's the concentration of A minus? That is, what's the concentration of acetate ion going to be here? [NOISE] How many moles of acetate did we dissolve in how many liters of water? One. One mole of sodium acetate, which means one mole of acetate ions dissolved in enough water to make one liter of solution. Our acetate concentration is going to be one molar. What's the acetic acid concentration going to be? 0.1 mole. 0.1 mole because we dissolve 0.1 mole of that in the same solution. Now what we have to do is get out the calculator and mess with the numbers,1 divided by 0.1 is simply 10, the LOG logarithm of 10 is 1, and so the final analysis, 4.74 plus 1, 5.74 for the pH of our acetic acid and acetate buffer solution. This is just using the Henderson-Hasselbalch equation. Again, the hardest part is usually figuring out which one is the acid, which was the conjugate base. Look for the extra H in the formula that'll tell you which one the conjugate acid is. Questions about this problem. Yes. Why don't you consider the sodium? Spectator ions. It floats around, does nothing, has no impact on pH. However, there is no such thing as a bottle of just acetate ions on the shelf by itself. There has to be some positive ion in there with it. Something like sodium acetate would be a reasonable place to get our acetate ions from [NOISE]. Any other questions about this slide? Anybody need more time with this slide? [NOISE]. Because now the second part of this problem, what happens when we take our buffer solution and do to it, what we did to our neutral water, that is put in 0.1 mole of HCl gas? Now remember when we did this to water, the pH dropped from seven to one, a factor of a million in terms of acidity. But buffer solutions are supposed to be able to resist changes in pH. Let's see if that happens [NOISE]. Here's our pH 5.74 buffer. We're going to add 0.1 mole of HCl gas to it. What happens? Let me just show you this one diagram that might make it clear qualitatively what happens [NOISE]. Suppose we have a buffer HX and X minus here, and then we add an acid. Well, acids generate H plus ions. What the H plus ions do is react with the X minus ions to generate more HX. What that means is the X minus concentration is going to go down by a little bit and the HX concentration is going to go up by a little bit. If you would have added a base, it would to be the other way around because the base is going to take a proton off of HX and make an X minus. This is what happens if you add a base. But in our situation we're adding an acid. The acetic acid concentration is going to go up a little bit, the acetate concentration is going to go down a little bit [NOISE]. Since HCl is a strong acid, we can assume that it reacts completely. That is, the 0.1 mole of HCl that we added reacts with 0.1 mole of acetate. Now, we originally had one mole of acetate in there so we convert 0.1 mole of it to something else, we still have 0.9 moles of it in there. The concentration of acetate decreases to 0.9 molar [NOISE]. Meantime, the acetic acid concentration is going to go up by the same amount because we had 0.1 mole of acetic acid in there originally, we're going to get 0.1 molar more when the 0.1 mole of HCl reacts with acetate ion. That brings that up to 0.2 moles [NOISE]. The rest of it, is just applying the Henderson Hasselbalch equation again. There's that equation. pKa hasn't changed, it's still 4.74. What has changed are the concentrations of acetate and acetic acid, which were 1 and 0.1 but are now 0.9 and 0.2. If we do the arithmetic, 0.9 divided by 0.2 is about 4.5. The logarithm of 4.5 is 0.65. Add that to 4.74, we get 5.39. [NOISE] Did the pH decrease? Yeah, a little bit. It was 5.74 now it's 5.39. Buffers don't necessarily keep pH as absolutely constant, but they do resist changes in pH. Remember that when we did the same thing to water with no buffer present, pH went down by six pH units. The solution became a million times more acidic. Here, it's going down by about 0.3 pH units. That's about a factor of two in terms of acidity. A factor of two more acidic versus a factor of a million more acidic, I would say the buffer solution did a pretty good job of resisting the change in pH [NOISE]. Do the concepts make sense? Because the arithmetic makes sense [NOISE]. We are almost out of time so I will simply say this about page 148 in the lecture notes [NOISE]. Title page 148 is Polyprotic acids and their salts. [NOISE] A polyprotic acid is an acid like carbonic acid, H_2CO_3 or phosphoric acid H_3PO_4, that has more than one proton that it can donate. Acids like that have several Ka values, Ka_1 Ka_2, and Ka_3. You don't have to copy all that now, it's in the lecture notes. The point is, a very valid way of making a buffer solution is to start with a polyprotic acid and then just titrate it until you have the pH that you want. You're doing titrations in lab this week, that's one thing they are good for [NOISE]. We'll stop there for today. [NOISE] Enjoy the weekend. [NOISE].We'll see you next time. [NOISE] I have a question [inaudible] [NOISE]. Okay, here's the point. These are times. Times and rates are not the same thing. Rate is inversely proportional to time. More importantly, there are BrF5 that can figure out a way through it. Now it's doing its thing in full point to seven minutes, the other gas is going in a shorter time, which means it's going faster. The point is, whatever this is, it should have a molecular weight that's less than BrF5. Now all of these have molecular weights less than BrF5 but the point is if you have these two upside-down, you're going to calculate a molecular weight that's greater than BrF5 or not, and that will match up with any of these guys. Okay. The point is there's a difference between time and rate. Rate A over rate B is the same thing as time B over time A. Okay. They are inversely proportional? Yeah. Just turn it upside down. All right. Thank you. [NOISE] I feel like this is a done question but obviously the brackets stand for concentration, right? The brackets stand for concentrations. The parentheses is just by way of separating Ka from Kb [NOISE]. Are we just using what we have? Okay. [OVERLAPPING] Take a look at the equations. All right. This is an equilibrium. We write the equilibrium products over reactants [NOISE]. These two on top, this one at the bottom and it will work. Okay. Same thing here. These two on top, this one at the bottom, [inaudible]. Okay. The point is this is the Ka expression, this is the Kb expression, it depends. Ka is this and Kb is that [NOISE]. I'm just asking how do you know what the average concentration? Okay. What I'm saying is that we don't know the numbers right now. Okay. But what I'm saying is when you do this algebraically, HA over HA cancels out. [NOISE] A minus, A minus cancels out. All you're left with is H_3O plus times OH minus and we know that the product of those two is just Kw. We know what the number for Kw is, it's 1 times 7 minus 14. Right. The point is we were down here, we have Ka for this, but we don't need Ka for that, we need Kb for this. But since those are conjugates to each other, we can use this which basically says Kb is equal to Kw divided by Ka , and then we do that. The point is if you take these two numbers and multiply them together, you get one times 7 to the minus 14. As long as you have one, you can find the other? Yeah, the point is if you know Ka for a weak acid, you can find Kb for its conjugate base and vice versa [NOISE]. I was confused because I wasn't sure if below this we were going to have the same thing.[OVERLAPPING]. I was just showing you where the expression came from algebraically. If it's a question for an exam, would you list all the molarities first or? Well, the point is there's x number of variables that we're going to have in any equation. If we have all of them except one you should be able to solve for the one or if it's an equilibrium technique, you can use an ICE table to calculate them. Okay. Yes. The point is the problems we've been solving in class are typical of the things that you're going to see on the final exam when we talk about this stuff. Is there a practice for final exam? There's nothing to practice final exam, but the thing I would say it is a multiple choice test. Okay. [NOISE] You've been taking multiple choice tests all semester long.
chem103-080-20191114-140000.mp4
From Dana Chatellier November 14, 2019
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