Buffers and Henderson-Hasselbalch | Chemistry | Khan Academy SAL: Let's say I have some weak acid. I'll call it HA. A is a place holder for really a whole set of elements that I could put there. It could be fluorine, it could be an ammonia molecule. If you add H it becomes ammonium. So this isn't any particular element I'm talking about. This is just kind of a general way of writing an acid. And let's say it's in equilibrium with, of course, and you've seen this multiple times, a proton. And all of this is in an aqueous solution. Between this proton jumping off of this and its conjugate base. A minus. And we also could have written a base equilibrium, where we say the conjugate base could disassociate, or it could essentially grab a hydrogen from the water and create OH. And we've done that multiple times. But that's not the point of this video. So let's just think a little bit about what would happen to this equilibrium if we were to stress it in some way. And you can already imagine that I'm about to touch on Le Chatelier's Principle, which essentially just says, look, if you stress an equilibrium in any way, the equilibrium moves in such a way to relieve that stress. So let's say that the stress that I apply to the system-- Let me do a different color. I'm going to add some strong base. That's too dark. I'm going to add some NaOH. And we know this is a strong base when you put it in a aqueous solultion, the sodium part just kind of disassociates, but the more important thing, you have all this OH in the solution, which wants to grab hydrogens away. So when you add this OH to the solution, what's going to happen for every mole that you add, not even just mole, for every molecule you add of this into the solution, it's going to eat up a molecule of hydrogen. Right? So for example, if you had 1 mole oh hydrogen molecules in your solution and you added 1 mole of sodium hydroxide to your solution, right when you do that, all this is going to react with all of that. And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen. And you can even play with it on the math. So this hydrogen goes down initially, and then it starts getting to equilibrium very fast. But this is going to go down. This is going to go up. And then this is going to go down less. Because sure, when you put the sodium hydroxide there, it just ate up all of the hydrogens. But then you have this-- you can kind of view as the spare hydrogen capacity here to produce hydrogens. And when these disappear, this weak base will disassociate more. The equilibrium we'll move more in this direction. So immediately, this will eat all of that. But then when the equilibrium moves in that direction, a lot of the hydrogen will be replaced. So if you think about what's happening, if I just threw this sodium hydroxide in water. So if I just did NaOH in an aqueous solution-- so that's just throwing it in water-- that disassociates completely into the sodium cation and hydroxide anion. So you all of a sudden immediately increase the quantity of OHs by essentially the number of moles of sodium hydroxide you're adding, and you'd immediately increase the pH, right? Remember. When you increase the amount of OH, you would decrease the pOH, right? And that's just because it's the negative log. So if you increase OH, you're decreasing pOH, and you're increasing pH. And just think OH-- you're making it more basic. And a high pH is also very basic. If you have a mole of this, you end up with a pH of 14. And if you had a strong acid, not a strong base, you would end up with a pH of 0. Hopefully you're getting a little bit familiar with that concept right now, but if it confuses you, just play around with the logs a little bit and you'll eventually get it. But just to get back to the point, if you just did this in water, you immediately get a super high pH because the OH concentration goes through the roof. But if you do it here-- if you apply the sodium hydroxide to this solution, the solution that contains a weak acid and it's conjugate base, the weak acid and its conjugate base, what happens? Sure, it immediately reacts with all of this hydrogen and eats it all up. And then you have this extras supply here that just keeps providing more and more hydrogens And it'll make up a lot of the loss. So essentially, the stress won't be as bad. And over here, you dramatically increase the pH when you just throw it on water. Here, you're going to increase the pH by a lot less. And in future videos, will actually do the math of how much less it's increasing the pH. But the way you could think about it is, this is kind of a shock absorber for pH. Even though you threw this strong based into this solution, it didn't increase the pH as much as you would have expected. And you can make it the other way. If I just wrote this exact same reaction as a basic reaction-- and remember, this is the same thing. So if I just wrote this as, A minus-- so I just wrote its conjugate base-- is in equilibrium with the conjugate base grabbing some water from the surrounding aqueous solution. Everything we're dealing with right now is an aqueous solution. And of course that water that it grabbed from is not going to be an OH. Remember, these are just equivalent reactions. Here, I'm writing it as an acidic reaction. Here, I'm writing it is a basic reaction. But they're equivalent. Now. If you were to add a strong acid to the solution, what would happen? So if I were to throw hydrogen chloride into this. Well hydrogen chloride, if you just throw it into straight up water without the solution, it would completely disassociate into a bunch of hydrogens and a bunch of chlorine anions. And it would immediately make it very acidic. You would get to a very low pH. If you had a mole of this-- if your concentration was 1 molar, then this will go to a pH of 0. But what happens if you at hydrochloric acid to this solution right here? This one that has this weak base and its conjugate weak acid? Well, all of these hydrogen protons that disassociate from the hydrochloric acid are all going to react with these OHs you have here. And they're just going to cancel each other out. They're just going to merge with these and turn into water and become part of the aqueous solution. So this, the OHs are going to go down initially, but then you have this reserve of weak base here. And Le Chatelier's Principle tells us. Look, if we have a stressor that is decreasing our overall concentration of OH, then the reaction is going to move in the direction that relieves that stress. So the reaction is going to go in that direction. So you're going to have more of our weak base turning into a weak acid and producing more OH. So the pH won't go down as much as you would expect if you just threw this in water. This is going to lower the pH, but then you have more OH that could be produced as this guy grabs more and more hydrogens from the water. So the way to think about it is it's kind of like a cushion or a spring in terms of what a strong acid or base could do to the solution. And that's why it's called a buffer. Because it provides a cushion on acidity. If you add a strong base to water, you immediately increase its pH. Or you decrease its acidity dramatically. But if you add a strong base to a buffer, because of Le Chatelier's Principle, essentially, you're not going to affect the pH as much. Same thing. If you add and acid to that same buffer, it's not going to affect the pH as much as you would have expected if you had thrown that acid in water because the equilibrium reaction can always kind of refill the amount of OH that you lost if you're adding acid, or it can refill the amount of hydrogen you lost if you're adding a base. And that's why it's called buffer. It provides a cushion. So it give some stability to the solution's pH. The definition of a buffer is just a solution of a weak acid in equilibrium with its conjugate weak base. That's what a buffer is, and it's called a buffer because it provides you this kind of cushion of pH. It's kind of a stress absorber, or a shock absorber for the acidity of a solution. Now, with that said, let's explore a little bit the math of a buffer, which is really just the math of a weak acid. So if we rewrite the equation again, so HA is in equilibrium. Everything's in an aqueous solution. With hydrogen and its conjugate base. We know that there's an equilibrium constant for this. We've done many videos on that. The equilibrium constant here is equal to the concentration of our hydrogen proton times the concentration of our conjugate base. When I say concentration, I'm talking molarity. Moles per liter divided by the concentration of our weak acid. Now. Let's solve for hydrogen concentration. Because what I want to do is I want to figure out a formula, and we'll call it the Hendersen-Hasselbalch Formula, which a lot of books want you to memorize, which I don't think you should. I think you should always just be able to go from this kind of basic assumption and get to it. But let's solve for the hydrogen so we can figure out a relationship between pH and all the other stuff that's in this formula. So, if we want to solve for hydrogen, we can multiply both sides by the reciprocal of this right here. And you get hydrogen concentration. Ka times-- I'm multiplying both sides times a reciprocal of that. So times the concentration of our weak acid divided by the concentration of our weak base is equal to our concentration of our hydrogen. Fair enough. Now. Let's take the negative log of both sides. So the negative log of all of that stuff, of your acidic equilibrium constant, times HA, our weak acid divided by our weak base, is equal to the negative log of our hydrogen concentration. Which is just our pH, right? Negative log of hydrogen concentration is-- that's the definition of pH. I'll write the p and the H in different colors. You know a p just means negative log. Minus log. That's all. Base 10. Let's see if we can simplify this any more. So our logarithmic properties. We know that when you take the log of something and you multiply it, that's the same thing as taking the log of this plus the log of that. So this can to be simplified to minus log of our Ka minus the log our weak acid concentration divided by its conjugate base concentration. Is equal to the pH. Now, this is just the pKa of our weak acid, which is just the negative log of its equilibrium constant. So this is just the pKa. And the minus log of HA over A. What we can do is we could make this a plus, and just take this to the minus 1 power. Right? That's just another logarithm property, and you can review the logarithm videos if that confused you. And this to the minus 1 power just means invert this. So we could say, plus the logarithm of our conjugate base concentration divided by the weak acid concentration is equal to the pH. And this right here, this is called the Hendersen-Hasselbalch Equation. And I really encourage you not to memorize it. Because if you do attempt to memorize it, within a few hours, you're going to forget whether this was a plus over here. You're going to forget this, and you're going to forget whether you put the A minus or the HA on the numerator or the demoninator, and if you forget that, it's fatal. The better thing is to just start from your base assumptions. And trust me. It took me a couple minutes to do it, but if you just do it really fast on paper-- you don't have to talk it through the way I did-- it'll take in no time at all to come to this equation. It's much better than memorizing it, and you won't forget it when you're 30 years old. But what's useful about this? Well, it immediately relates pH to our pKa, and this is a constant, right, for an equilibrium? Plus the log of the ratios between the acid and the conjugate base. So the more conjugate base I have, and the less acid I have, the more my pH is going to increase. Right? If this goes up and this is going down, my pH is going to increase. Which makes sense because I have more base in the solution. And if I have the inverse of that, might be just going--