Chemistry: Weak Acid-Strong Base Titration

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When we consider titrating a weak acid with a base, we have a very different situation than titrating a strong acid. We're still interested in the total number of protons, but, unlike a strong acid, the weak acid is holding most of those protons and it's not until we actually do the process of titration that we start to remove them. So the pH is going to behave very differently. It's going to respond in a very different way than what we saw with a strong acid. So let's first familiarize ourselves with a titration curve for a weak acid, and then we'll come back to the issue of what this is good for or why anyone would want to titrate a weak acid.
As an example, let's suppose that we had 100 milliliters of a 10^th molar solution of acetic acid, of vinegar. And we're going to titrate, again, with a base that's sufficiently strong, that it can easily pull the protons off of this weak acid. In this case, 10^th molar sodium hydroxide solution will do the trick. Now, once again, we're going to look at pH as a function of how much hydroxide we add to the solution. First of all, we know that, as we add hydroxide, we've got to increase the pH throughout the entire process, so that's a given. We're adding, again, more base, so the pH is going to have to increase, reflecting the fact that the total H^+ concentration is decreasing. But, because this is a weak acid, our pH initially is not going to be 1, like it was in the case of HCl, but it's going to be somewhat higher, because, again, this is a weak acid, not a strong acid. How would actually calculate this point? Well, this is just a K[a] problem at this point, because we have a weak acid dissolved in water, we know how to convert that to H^+ concentration, using the acid dissociation equilibrium and that, in turn, gives us H^+ and from there we get pH. That would turn out to be, if we do that calculation, that we would get a pH of 2.9 for that initial solution of 10^th molar acetic acid.
As we start to add base, think about what's happening. As we add base, we're consuming some of that acetic acid. And let's consider now this point, where we get to consuming half of it. At this stage, we have half an equivalent of acetic acid still, in other words, half as much as we started with, but we also have an equal amount of the conjugate base, in other words, of the acetate ion. Now, let's go back and think about this. If we have equal amounts of a weak acid and its conjugate base, we have a buffer system. We have an equation that describes a buffer system and what the pH should be for that buffer system, so let's review quickly the Henderson-Hasselbalch equation, which tells us that pH is equal to the pK[a] plus the log of this ratio. Well, if these two have equal concentrations, this entire term drops out. Once again, the log of 1 is going to be zero, and so the pH will be equal to the pK[a]. So, at that half-equivalence point, we have the unique position that the pH is exactly equal to the pK[a]. The pK[a] of acetic acid is 4.74, so that would be what the pH is right at this point. But, because it's a buffer system, we know that buffer systems respond very, very slowly to increase concentration of hydroxide or of H^+, and so the pH we'd expect not to change very much as we continue to add hydroxide. Indeed, this entire region throughout here is called the buffer region, because it's a buffered system. We have both acid and base present, and so, as we know from studying buffers, there's very little change in the pH through that region. It's not until we approach the buffer capacity, where we start to consume the last bits of acetic acid, where the pH goes through a radical change. So, once again, our pH becomes unstable, it starts to take off and it goes through an equivalence point, just like what we saw with a strong acid. At the equivalent point, again, that's the point at which we've added 1 equivalent of hydroxide, in other words, exactly enough hydroxide to completely neutralize the acid and no more.
Unlike a strong acid, the equivalence point does not show up at a pH of 7 when we're titrating a weak acid. Why would that be? Well again, let's ask what is in solution at that point. We've consumed all the acetic acid now with the hydroxide. The only thing left at this point is acetate, sodium acetate in this case. Well, what would be the pH of a sodium acetate solution? We know that it's not going to be 7. We know acetate is basic, so the pH would be above 7. So how would we actually calculate with the pH is at the equivalence point? Well, if we knew the concentration of acetate, then we'd be able to use a K[b] expression, and that would tell us then concentration of hydroxide. Once we knew the concentration of hydroxide we could convert that to H^+ and from there, figure out the pH. So let's go ahead and do that to get a little more experience. Again, the question that we're going to ask is, "What is the pH at the equivalence point for this titration curve of acetic acid?"
We've added, in this case, 100 milliliters of sodium hydroxide. We started out with 100 milliliters of acetic acid, so our volume has doubled, and so our concentration of acetic acid, which is .1 moles per liter, that actually will now be cut down to half that amount. In other words, the moles of acetic acid we started with, no matter what it is, that's now been converted to moles of acetate, but the volume has doubled. So, in fact, our concentration of acetate will be .01 moles per liter divided by 2, because we're doubled the total volume in this particular problem. I'll go ahead and do that calculation, just in case you need to do this type of a problem with different volumes. At this stage we'd know the acetate concentration, that will allow us to figure out hydroxide and from there we get pH. So let's go ahead and do that.
Okay, now, we need the information that K[b] for acetic acid is 5.6 x 10^-10. And, of course, if we didn't have that information, we'd get that from K[a]. Now, we need to know what our number of moles of acetic acid is, and we'll then convert that to moles of acetate. So we know that we have .1-liter times our concentration. And, just a quick reminder here, chemists often abbreviate the acetate ion as OA[c]. So this is just a conjugate base of acetic acid and HOA[c] is acetic acid itself. So now, if we have .010 moles of acetic acid, we have .010 moles of acetate now, after we've consumed that amount. So again, we're adding enough base to completely consume the acetic acid and we now have an equivalent amount of acetate. So we have 0.010 moles now of acetate that we got from consuming the acetic acid divided by the volume total, which is 200 milliliters - remember we titrated with 100 milliliters, we started with 100 milliliters, so .200 liters, and that gives us then a concentration of 0.05 molar of acetate ion. Okay, reality check here, remember that we started out with 10^th molar acetic acid, we've doubled the volume now, so our concentration should be half as much as it was of acetic acid. So half of .1 is .05. So that's how much acetate we have. Now we have our K[b] problem. We've got acetate, we're going to use K[b ].6 x 10^-10, and that's going to be equal concentration of hydroxide, which we don't know, and concentration of acetic acid, which we're going to assume is equal to the amount of hydroxide that we get. Notice a crucial assumption is that the hydroxide all comes from acetate reacting with water to give acetic acid. So again, we'll assume that those are the same values, and we'll have to check that, divide by the amount of acetate we have, which is going to be at equilibrium .05 minus whatever combined with water. And we're going to assume also here that x is going to be small compared to.05. We hope that x is big enough that it's the main source of hydroxide, but small enough that the acetate we have at equilibrium is essentially equal to the acetate that we had initially.
So converting this then, and I'll draw it this way, simply rearranging and solving for x, where I'm going to simplify my math, assume that this is .05 now. I'm going to get a value for x - remember that's hydroxide concentration of 5.3 times 10^-6 moles per liter. So reality check again, did our assumption work? This is significantly larger than 10^-7. It's getting close, but we're still okay, we're within 2 to 3 percent, and it's significantly small compared to .05. So our assumptions work, our simplifying assumptions make the math easier. If we convert that hydroxide concentration to concentration of H^+ through using K[w] and then convert that to pH, we end up at the end of the day with a pH equal to 8.72. Notice the pH is more basic than it would be if we had the equivalence point for a strong acid. And again, that makes sense. It has to be more acidic than for a strong acid, because we're left with the conjugate base, and that's going to cause our pH again to be above 7 in this case. I should clarify that a little bit: we get the conjugate base with a strong acid as well, but the conjugate base with a strong acid is sufficiently weak that it doesn't change the pH from 7. In this case, because it was a weak acid, the base is stronger, although it's still weak, and that raises our pH above 7.
So, summarizing then, we can calculate a pH initially with a K[a ]calculation, in the middle with Henderson-Hasselbalch, this is just a buffer problem, because we have both acid and base. And then, at the end, doing a K[b] calculation to find the pH so we understand how to get - in fact, at any point along this curve, in the buffer region, we know how to treat, because it's just figuring out the amount of acetate and the amount of acetic acid simply by doing a neutralization reaction, like we did for the example of half an equivalent. And so, the really useful thing about a weak titration curve is that if we do have an unknown acid, in this case we knew it was acetic acid, we could titrate, find the equivalence point and from that figure out ultimately the molecular weight of the acid, like we described previously. But also, by taking half he equivalence point, we can figure out what the pH is, and that corresponds to the pK[a]. So for nothing extra, we get a good value for the acid strength of the unknown acid. So, in fact, that's one of the main uses of a titration curve for an unknown acid, is very accurately measuring just how strong the acid is. So we've looked at a strong acid, we've looked at a weak acid, our next step is going to be to consider what a titration curve looks like for an acid that's capable of giving up more than 1 proton. And we'll look at that next.
Equilibrium in Acqueous Solution
Acid-Base Titration
Weak Acid - Strong Acid Titration Page [2 of 2]