In this lecture, we're going to continue our study of these neutralization reactions in which we mix together an acid and a base and examine what the properties of the solution are. Remember up to this point this is what we've discussed. We'll take the base B and the acid HA, and we'll mix them together. And we've always assumed up to this point equal moles of acid and base. Under those circumstances what we've observed is if we meant to get, mix together equal numbers of moles of strong acid and strong base then the solution is in fact neutralized. But if we mix together equal moles of strong base with weak acid, the solution is basic. Or equal moles of strong acid with weak base, the solution is acidic. What we'd now like to do is actually go back and test the consequences of not assuming that there are equal moles of acid and base. So here's an example, based upon what we were looking at before. We're going to mix together nitrous acid and sodium hydroxide, just like we did before. But we're going to make, mix unequal numbers of moles of these two substances. Let's count the number of moles of the, nitrous acid that we began with. It is equal to, let's see, we have taken 0.5 liters and multiplied by 0.10 molar, so we have .05 moles of the nitrous acid. If we take the number of moles of the ammonia, I'm sorry, of sodium hydroxide, it, we've taken 0.25 liters, multiplied by 0.10 moles gives us a total number of moles of 0.025 moles. If we compare these two then we put in more moles of a weaker acid and fewer moles of a stronger base. So it's reasonable to ask the question, essentially, which one of those wins. Is it more important that the base is strong or is it more important that we had put in more moles of the acid. So, is this solution going to be acidic, basic, or neutral, when we carry out this particular reaction between the hydroxide and the nitrous acid. Remember when we discussed, much earlier, the idea of a limiting reactant? And which one of these we're going to run out of first? Clearly in this particular case, when the reaction takes place. We're going to run out of the sodium hydroxide before we run out of the nitrous acid. Consequently, it is the limiting reactant and looking at the reaction above, the number of moles of nitrous an-ion that we produced is clearly limited by the amount of sodium hydroxide we put in to the solution, and therefore is equal to 0.025 moles. And, correspondingly, the number of moles of nitrous acid still in the solution is the amount of nitrous acid which reacted, which is equal to the amount of sodium hydroxide in the solution subtracted from the amount we started off with. So it would be equal to 0.05 moles minus 0.025 moles which turns out to be equal to 0.025 moles as well. Now. We need to be careful in calculating the concentrations of these materials in solution, so when we do this, we'll calculate the nitrous i, anion concentration as being 0.025 moles divided by the volume of the solution. Let's check the volume of the solution. Is going to be 750 milliliters because we've mixed together two solutions. So that's 0.750 liters. And if we take that and just plug it into the calculator, we get a concentration of 0.033 molar. And likewise the concentration of the nitrous acid in solution. Is equal to 0.033 molar. So I have a particularly interesting solution here in which we have, in the solution a combination of the acid, and its conjugate base mixed into the solution at the same time. So, what are the properties of such a solution? And it turned out to be very interesting. Let's try to determine now what we think the pH of this particular solution will be so we can answer the question, whether it's acidic base or neutral and then test that hypothesis. What we'll then do is take our usual reaction, nitrous acid plus water. Gives us the hydroxide ion concentration plus the hydroxide plus the nitrous anion. That's our reaction in our Rice table. Our initial conditions now, before we've come to equilibrium are, that the concentration of the nitrous acid is 0.033. The concentration of the nitrous an-ion is 0.033. And we'll approximate that the concentration of the hydronium ion, for now, is zero. As we change towards equilibrium, we'll assume that we're going to shift in the amount of the nitrous acid towards the nitrous an-ion by minus x. That will produce x moles per liter of hydronium. And X moles per liter of the nitrous anion, and we'll reach equilibrium with concentrations 0.033 minus x, x, and 0.033 plus x. We now put those into our acid ionization equilibrium constant to give us an equation which we can solve to find the value of x, which is the hydronium ion concentration. We'll remember from a previous slide that the value of this equilibrium constant is 5 times 10 to the minus 4. If that's the case, then we now have 0.033 plus x, multiplied by x, divided by 0.033 minus x, is equal to 5 times 10 to the minus 4. And now we have an equation that we need to solve for x. To determine what the value of the hydronium ion concentration is going to be. We could go through the trouble to write this out as a quadratic equation, but we could actually take an assumption here, in which x is assumed to be a small number. For example, if x is quite a bit smaller than 0.033, and that's something we can test in a moment when we solve the equation, then 0.033 plus x is really just the same as 0.033 to a good approximation. And 0.33 minus x is also equal to x. To that same approximation. Correspondingly, we ought to be able to then write, as a good approximation, that Ka is now equal to 0.033 times x, divided by 0.033, and that should be equal to 5 times 10 to the minus 4. And if that's the case, then x is itself equal to 5 times 10 to the minus 4. Notice that 5 times 10 to the minus 4 is in fact dramatically smaller than 0.033, very substantially smaller. Consequently, this is a good approximation. And that means that this is a good approximation to the answer to the equation. We can use this value of x because it's equal to the hydronium ion concentration, according to our table up above, to calculate what the pH of this particular solution is. And the pH of the solution then turns out to be equal to 3.3. So we now, we have an answer to our question. The solution is in fact acidic. That it is more important that we have excess acid in the solution, because actually the base has been essentially fully neutralized. There is no hydroxide still floating around from the sodium hydroxide. There's a little hydroxide left in there from the auto ionization of water. But overall, what we primarily have are the acid and its conjugate base. Nitrous acid is the stronger acid than nitrous anion is a base, and so the solution turns out to be acidic. This kind of solution actually goes by a special name, and has particular properties that we turn out to be interested in. They are referred to as a buffer solution. A buffer solution is a solution in which we have an acid and its conjugate base in solution together. Those turn out to have the very interesting property. That they have relatively constant pH, even when we add a strong ace, a strong acid, or a strong base to it. For example, if I were to modify this particular problem, by instead of having at the outset taken 0.50 liters of the nitrous acid. And 0.025 liters of the sodium hydroxide. Imagine we dumped in another 50 milliliters of the sodium hydroxide. That's quite a bit of additional base. That would change this volume here to 0.30 liters. Change the number of moles of sodium hydroxide to 0.30. That would also change the amount of nitrous nitrous acid produced to 0.030. It would change the amount of nitrous acid still available to 0.020. We could go down here and actually modify the entries in the table quite quickly. But in the end, what we would discover is that, in this modified solution, if we were to add in, instead, 50 milliliters of 0.1 molar solution, it turns out in that particular case, the pH becomes 3.4. Whereas just a little while ago, sorry, it's actually 3.5. Whereas a little while ago we determined that the solution actually has pH 3.3. The addition of 50 milliliters of concentrate, of fairly concentrated sodium hydroxide solution, has only slightly changed the pH of the solution. That's a property of a buffer solution, and it's in fact what gives them their names. That apparently changes in pH are buffered in these solutions by the presence of both the acid and the base at the same time. Why would this be true? Well, one way to understand that is actually to consider it in terms of Le Chatelier's principle. Let's rewrite our reaction here again, HNO2 plus H2O, goes to NO2 minus, plus H3O plus. Oops, not on the screen. If I were to add strong acid to this particular reaction. That strong acid would react with the nitrous anion and be consumed by it, neutralized by it. Removing some of this material from the solution will cause the HNO2 to shift back over, creating some additional hydronium ion, but buffered against it to a certain extent by just the fact that as we have added say, the extra acid, the solution responds to try to consume that acid. Or if I were to add base to the solution it would neutralize a certain amount of the HNO2 causing the reaction to shift from right to left, consuming some of the hydronium but balancing those consumptions. In the way that Le Chatelier's principle says that we will always respond to shift the stresses away from the reactant the change that we have made as we go to equilibrium. There's an equation which is often applied to this based upon the approximation we've made just a little while ago. Let's take the general expression, Ka, which is the anion concentration times the hydronium ion concentration divided by the acid concentration. Let's take the logarithm base ten of that on both sides and in the process of taking the logarithm base ten I'm going to go ahead and split up the expression here in to two terms, using the properties of logarithms. This is exactly the same equation as the equation above, we've just used the properties of logarithms. Notice here though, I could also now multiply through by a minus sign, if I wanted to. And in doing so, I've now actually generated, on this side, clearly, the pH, and over here, we're actually going to define a new term that we'll call the PKa. Clearly, the PKa is just the negative logarithm base ten of the equilibrium constant itself. The relationship between pH and the PKa of the acid is, according to the equilibrium constant expression, minus the logarithm of the an-ion, divided by the concentration of the acid. Now, you'll recall when we did the calculation a little while ago. I'm going to bring it back up on to the project, on to the the camera here. That we made the approximation, that the concentration of the acid is about the same at equilibrium as it was when we started off. Because we assumed that x was tiny, compared to the initial concentration of the acid. And it turned out that it was, x was quite small as we looked down below. And likewise, the concentration of the anion we assumed was the same at equilibrium as it was initially, because x is quite small. If that's the case, then we might take each of the concentrations here of the anion in the acid and approximate that they are the same at equilibrium, as they were initially. The expression we've written right now is exact. We're now going to make an approximation to that equation by replacing the concentration of the anion with its initial concentration. And the concentration of the acid with its initial concentration. Now let's just move this term to the other side of the equation. The pH of a solution, of a buffer solution, is equal to the PKa of the con, of the weak acid, which is in that solution plus the logarithm base ten of the concentration of the anion that we began with divided by the concentration of the acid that we began with. This particular equation, goes by a name. It's called the Henderson-Hasselbach equation. I've prepared it up here on the screen so that hopefully it's a little bit easier to read. But notice one of the consequences of this equation is that first. If the anion concentration of the acid concentration are the same, then this ratio is 1. The logarithm base 10 of 1 is 0, and the pH will become equal to the PKa. In other words, the pH of a buffer solution in which we have equal amounts of the acids and base can be predicted just by looking at the PKa. Secondly, any changes that we make in either A minus or HA result not in changes in this ratio in the pH but rather only in the logarithm. And as a consequence, the pH will vary only slightly. As we say, add some acid, causing the base to shift to acid, or add some base, causing the base the acid to shift to base. Then in either of those circumstances, the pH will remain relatively constant. Buffer solutions have wide applicability in chemistry, most notably in biochemistry and biomolecular systems. Where balancing the pH and maintaining a relatively constant pH can be important for biophysical processes, for example, and this lecture helps explain why buffers have the property that they do.