Buffer solutions

- [Voiceover] Buffer solutions resist changes in pH and so let's think about a solution of a weak acid and its conjugate base. So here we have HA which is our generic weak acid and so the conjugate base would be A-. A buffer solution needs to have substantial amounts of both present and that's what I'm trying to represent over here. We have a beaker that has in this case equal amounts of HA and A-. To our buffer solution, we're gonna add some strong acids so a source of protons. I'm going to draw a proton in here so H+. I could have written H3O+ because H+ and H2O give you H3O+. I'm just writing H+ to make it a little bit easier to think about. If you're increasing the concentration of H+ ions, or you're increasing the concentration of hydrionium ions in solution, you would think that would decrease the pH dramatically and that would happen if you didn't have something to react with the protons in solution. We do in our buffer solution, right? We have A-. We have our base. The base is going to pick up this proton and H+ and A- are going to form HA. We're going to make more of our weak acid here. Let's see, we started out with some weak acids. Let me go ahead and draw that in here so some HA and then we're going to make more because H+ and A- give us HA so we're making more of our weak acid here. What about A-? We're gonna use one of these up to make HA so now we have only two of these left. We're decreasing the concentration of A- but what we've done is we've effectively removed protons from solution and that's how a buffer is able to resist a change in pH if you add acid. What about if you add base? Next, let's think about our buffer solution and this time we're going to add a strong base. We're going to increase the concentration of hydroxide ions in solution. If we increase the concentration of hydroxide ions in solution, you would think that would increase the pH. It would except for the fact that we have an acid that can react with our base. We have some weak acid present here. The hydroxide ion is going to pick up a proton from HA so we have H+ here. OH- and H+ give us H2O so we're going to make water here. We're going to lose some of our weak acid. We started with three and now we have only two left here. If OH-, if hydroxide takes a proton away from HA, then we're left with A-. We're going to increase the amount of A- that we have in solutions. Let me go ahead and draw that in here. So once again we've effectively buffered against a change in pH because the hydroxide ions that we added reacted with the acid that was present. We've effectively removed hydroxide ions from solution. So that's the idea of a buffer solution. Let's see if we can derive an equation that will allow us to do some calculations. Next let's look at what we have down here. We know that we have in a buffer we have substantial amounts of HA and A- present. We have this reaction at equilibrium. For our equilibrium expression, concentration of products over concentration of reactions. Here we have concentration of products over our concentration of reactions once again leaving water out. Let's take the negative log of both sides of this. We're gonna take the negative log of Ka so that would be -log of Ka here and the negative log of all of this which would be equal to the negative log of all of that. I'm going to write it a little bit differently. I'm gonna put the concentration of hydronium ions out front here and then have concentration of A- over concentration of HA. I wrote it like this because it makes it a little bit easier to see a property of a logarithm. Let me just go ahead and write down the property that I'm referring to. If you have log of A times B, that's the same thing as log of A plus log of B. Here we have the negative log so the negative log of AB is equal to the negative log of A plus the negative log of B. I'm just going to write minus the log of B here. In this case, concentration of H3O plus that would be A so this would be A. This over here would be B. Let's think about what that would give us now. Let's get some more space and let's think about our log properties so on the left side we have negative log of Ka. On the right side, we would have negative log of A so that would be negative log of concentration of H3O+ minus the log of B here so that would be minus the log and B was this right here. Minus the log of A- over HA. We can keep going here because we know the negative log of KA is equal to the PKa so this is equal to the PKa. Over here we have the negative log of the concentration of hydronium ions that's equal to the pH so negative log of concentration of H3O+ is equal to the ph and then we have minus log of the concentration of A- over the concentration of HA. We can re-write this for our final equation so we're just gonna write this all for pH. The pH is equal to PKa + the log of A- over the concentration of HA here and this is called the Henderson-Hasselbalch equation. Right here is the Henderson-Hasselbalch equation. It's very useful when your doing buffer calculations. We'll look at examples of this in the next video.