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# pKa and pKb relationship

Video transcript
In the last video we learned that if I had some -- let's say I have some weak acid. So it's hydrogen plus some-- the rest of whatever the molecule was called. We'll just call it A. And I think this tends to be the standard convention for the rest of the acid. They can disassociate, or it's in equilibrium because it's a weak acid. So it can be in equilibrium with -- since it's a weak acid, it's going to produce some hydrogen proton. And then the rest of the molecule is going to keep the electrons. So it's going to be plus -- oh, this is in an aqueous solution. Let me do that. Aqueous solution. And then you're going to have the rest of the acid, whatever it might be. A minus, and that's also going to be in an aqueous solution. And that's the general pattern. We've seen the case where A could be an NH3, right? If A is an NH3, then when you have this, you have an NH4 plus, and this would be ammonia. And this is just NH3. A could be a fluorine molecule right there, because then this would be hydrogen fluoride or hydrofluoric acid. And this would just be the negative ion of fluorine. Or a fluorine with an extra electron. So it could be a bunch of stuff. You can just throw in anything there, and it'll work. Especially for the weak acids. So we learned a last video that if this is the acid, then this is the conjugate base. And we could write the same reactions, essentially, as kind of more of a basic reaction. So we could say, if I start with A minus -- it's in an aqueous solution -- that's in equilibrium with-- This thing could grab a hydrogen from the surrounding water and become neutral then. It's still in an aqueous solution. And then one of those water molecules that it plucked that hydrogen off of is now going to be a hydroxide molecule. Right? Because it's hydrogen. Remember, whenever I say pluck the hydrogen, just the proton, not the electron for the hydrogen. So the electron stays on that water molecule, so it has a negative charge. It's in an aqueous solution. So we could write the same reaction both ways. And we can write equilibrium constants for both of these reactions. So let's do that. Let me erase this, just because I can erase this stuff right there, and then use that space. So an equilibrium reaction for this first one. I could call this the K sub a, because the equilibrium reaction for an acid. And so this is going to be equal to its products. So the concentration of my hydrogen times my concentration of whatever my conjugate base was, divided by my concentration of my original acid. My weak acid. So this would be the concentration of HA. Fair enough. I could also write an equilibrium constant for this basic reaction. Let me do it right down here. So I'll call that my K sub b. This is a base equilibrium. And so this is equal to the concentration of the products. It's becoming tedious to keep switching colors. Actually, I'll do it. Because it makes it easier look at, at least for me. HA times the concentration of my hydroxide ions divided by my concentration of my weak base. A minus. Remember, this can only be true of a weak base or a weak acid. If we were dealing with a strong acid or is or a strong base, this would not be an equilibrium reaction. It would only go in one direction. And when it only goes in one direction, writing this type of equilibrium reaction makes no sense -- or equilibrium constant-- because it's not in equilibrium. It only goes in one direction. If A was chlorine, if this was hydrochloric acid, you couldn't do this. You would just say look, if you have a mole of this, you're just dumping a mole of hydrogen protons in that solution and then a bunch of chlorine anions who are not going to do anything. Even though they are the conjugate base, they wouldn't do anything. So you can only do this, remember, for weak acids and bases. So with that said, let's see if we can find a relationship between Ka and Kb. What do we have here? We have an A minus on both sides of this. We have H over-- OH over A minus. Let's solve for A minus. Right? If we multiply both sides of this equation on the left-hand side we get Ka times the inverse of this. So you have your HA over H plus is equal to your concentration of your conjugate base. A minus. And let's do the same thing here. Solve for A minus. So to solve for A minus here, we might have to do 2 steps. So if we take the inverse of both sides, you get 1 over Kb is equal to A minus over H, the concentration of my conjugate acid times the concentration of hydroxide. Multiply both sides by this. And I get A minus is equal to my concentration of my conjugate acid times concentration of hydroxide. All of that over my base equilibrium constant. Now, these are the same reactions. Right? In either reaction for given concentrations, I'm going to end up with the same concentration. This is going to equal that. Right? These are two different ways of writing the exact same reaction. So let's set them equal to each other. So let me copy and paste it, actually. So I'm saying that this thing, copy, is equal to this thing right here. So this is equal to-- let me copy and paste this-- that. That's equal to that. So let's see if we can find a relationship between Ka and Kb. Well, one thing we can do is we can divide both sides by HA. Right? So if we divide both sides by HA. Actually, I could probably have that earlier on to the whole thing. If we ignore this part right here, this is equal to that. Let me erase all of this. Oh. I'm using the wrong tool. So we could say that they both equal the concentration of A minus. So that's equal to that. We can divide both sides by HA. This over here will cancel with this over here. And we're getting pretty close to a neat relationship. And so we get Ka over our hydrogen proton concentration is equal to our hydroxide concentration divided by Kb. You can just cross-multiply this. So we get Ka, our acidic equilibrium concentration, times Kb is equal to our hydrogen concentration times our hydroxide concentration. Remember, this is all in an aqueous solution. What do we know about this? What do we know about our hydrogen times our hydroxide concentration in an aqueous solution? For example, let me review just to make sure I'm jogging your memory properly. We could have H2O. It can autoionize into H plus. Plus OH minus. And this has an equilibrium. You just put the products. So the concentration of the hydrogen protons times the concentration of the hydroxide ions. And you don't divide by this because it's the solvent. And we already figured out what this was. If we have just completely neutral water, this is 10 to the minus 7. And this is 10 to the minus 7. So this is equal to 10 to the minus 14. Now, these two things could change. I can add more hydrogen, I could add more hydroxide. And everything we've talked about so far, that's what we've been doing. That's what acids and bases do. They either increase this or they increase that. But the fact that this is an equilibrium constant means that, look, I don't care what you do to this. At the end of the day, this will adjust for your new reality of hydrogen protons. And this will always be a constant. As long as we're in an aqueous solution, a solution of water where water is a solvent at 25 degrees. I mean, in just pure water it's 10 to the minus 7. But no matter what we do to this and this in an aqueous solution, the product is always going to be So that's the answer to this question. This is always going to be 10 to the minus 14. If you multiply hydrogen concentration times OH concentration. Now they won't each be 10 to the minus 7 anymore, because we're dealing with a weak acid or a weak base. So they're actually going to change these things. But when you multiply them, you're still going to get 10 to the minus 14. And let's just take the minus log of both sides of that. Let me erase all this stuff I did down here. I'll need the space. Let's say we take the minus logs of both sides of this equation. So you get the-- let me do a different color-- minus log, of course it's base 10, of Ka. Let me do it in the colors. Ka times Kb is going to be equal to So what is this equal to? The log of 10 to the minus 14 is minus 14, because 10 to minus 14th power is equal to You take the negative of that, so this becomes 14. So the right-hand side of your equation just becomes 14. And this one, we could use log properties. This is same thing as the minus log of Ka. We use the colors. Ka... plus the minus log of Kb. Or, though you could think of this-- this is your pKa, this is your pKb. So you can say, this is pKa plus pKb. Oh, I wanted to use blue. Plus pKb, and all of that's going to be equal to 14. Now why is this useful? Well, if you know the pKa for a weak acid-- For example, let's say we have NH4 plus. This is a weak acid, right? It can donate an H, but it's not an irreversible reaction. That H can be gained back. So this is a weak acid. If you look it up on Wikipedia, it'll say, hey, the pKa of NH4 is equal to 9.25. Right? So this is 9.25 for NH4. For ammonium. So what is going to be the pKb for ammonia? Right? Let me write that reaction down. So this is NH4. Is in equilibrium. This is plus. It can get rid of one of its hydrogen protons, and you're just left with ammonia. So this is the acidic reaction. So this is what the pKa is associated with. So the equilibrium constant for this reaction is is equal to 9.25. And if I had the reverse reaction, the conjugate base reaction, so ammonia converts to ammonium. Plus it grabbed that hydrogen proton from a water molecule. If I wanted to figure out the pKb, or the equilibrium constant, or the negative log of the equilibrium constant for this reaction, what is it? Well, this one's 9.25. And 9.25 plus this pKb have to be equal to 14. So what's 14 minus 9.25? It's what, 4.75. So we immediately know the equilibrium constant for the conjugate base reaction. So it's a useful thing to know. That the pKa plus the pKb is equal to 14. And always remember, you see these pKa and pKb, and you say, what is that? Well, if you see a p, it's a negative log of something. And in this case, it's the negative log of the equilibrium constant for an acidic reaction. Plus the negative log of the equilibrium basic reaction, where this is the conjugate base of this acid. It's always going to be equal to 14 if we're dealing with an aqueous solution at 25 degrees Celsius, which is essentially room temperature, which is usually going to be the case.