Titrating a polyprotic acid with a strong base produces a pH curve with as many equivalence points as there are acidic protons on the acid. The pKₐ values for these protons can be estimated from the corresponding half-equivalence points on the curve, where pH = pKₐ. Created by Jay.
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- What would the pH values be at the equivalence points?(1 vote)
- The actual value of the equivalence point (the pH at least) depends on the identity of the analyte and the titrant, so it’ll depend on the actual titration reaction. But the equivalence point occurs when the moles of an acid equal and neutralize the moles of a base. On a titration curve graph this happens at the inflection points, or the middle of the vertical portions of the graph.
In this reaction since it’s a diprotic acid, we’ll have two equivalence points. We can estimate from the graph by looking at the corresponding pH values from the inflections points. The first equivalence point at3:20corresponds roughly to a pH of 6. The second equivalence point at6:20corresponds roughly to a pH of 12.
Hope that helps.(1 vote)
- Why would the pH rise between the one equivalence point and the second equivalence point as the newly introduced OH- interacts with HA?(1 vote)
- The difference in the equivalence points for a polyprotic acid is due to the different levels of acidity of the protons. The first equivalence point is reached when enough base has been added to neutralize the first acidic proton. In this instance, H2A becomes its conjugate base HA^(-) at the first equivalence point. The second acidic proton is less acidic than the first and less reluctant to be removed in an acid-base reaction so it only does so in a more basic environment, or at a higher pH. So at the second equivalence point HA^(-) becomes its conjugate base A^(2-).
This is generally true for polyprotic acids that the remaining acidic protons become less acidic as successively more protons are removed and require a more basic pH to come off.
Hope that helps.(1 vote)
- [Instructor] A polyprotic acid is an acid with more than one proton that it can donate in solution. An example of a polyprotic acid is the protonated form of the amino acid alanine. Here's a dot structure showing the protonated form of the amino acid alanine, and we can represent this as H2A plus. Let's say we're doing a titration with the protonated form of alanine, and we're adding sodium hydroxide to an aqueous solution of the protonated form. The protonated form of alanine has two acidic protons. So one acidic proton is on the oxygen, and the other acidic proton is one of the protons bonded to the nitrogen. The proton bonded to the oxygen is the more acidic of the two. Therefore, when we add some hydroxide anions to the solution, the hydroxide anions will pick up this proton and form water and convert the protonated form of alanine into the overall neutral form of alanine, HA. Let's say that we start out in our titration with one mole of the protonated form of alanine, and then we add in our hydroxide anions. After we've added in one mole of hydroxide anions, we've completely neutralized the protonated form of alanine, and H2A plus is converted into HA. So one mole of H2A plus reacts with one mole of hydroxide anions to form one mole of HA. So now we have one mole of HA in solution. And if we continue to add hydroxide anions, hydroxide anions react with HA and take one of the acidic protons on the nitrogen to form A minus. A minus is the conjugate base to HA. It would take another mole of hydroxide anions to completely neutralize the HA and to turn it into A minus. So one mole of HA reacts with one mole of hydroxide anions to form one mole of A minus. So going back to the protonated form of the amino acid alanine, since the protonated form has two acidic protons, we call this a diprotic acid. So let me go ahead and write in here diprotic. And because there are two acidic protons on the protonated form of alanine, there will be two pKa values, which we can figure out from the titration curve. Next, let's look at the titration curve for the titration of our diprotic acid with sodium hydroxide. On the y-axis is pH, and on the x-axis is moles of hydroxide anions added. Before we've added any hydroxide anion, so this point right here on our titration curve, we have our diprotic acid present, so H2A plus. And since we started with one mole of H2A plus, it would take one mole of hydroxide anions to completely neutralize the H2A plus. So after one mole of hydroxide anions has been added, that brings us to equivalence point one. So we go to one mole of hydroxide anions on the x-axis, and we'd go up to where that intersects our titration curve. And this point on our titration curve is equivalence point one. At equivalence point one, all of the H2A plus has been converted into HA. So now we have one mole of HA present. Next, let's think about adding 0.5 moles of hydroxide anions. So that would only neutralize half of the H2A plus that was initially present. Therefore, if we go to 0.5 moles of hydroxide anions and we'd go up to where that intersects our titration curve, this point represents the half equivalence point or half equivalence point one for this titration. Next, let's look at half equivalence point one in more detail. So here is half equivalence point one on the titration curve. So if we started out with one mole of H2A plus and we've added 0.5 moles of hydroxide anions, we've neutralized half, or 0.5 moles, of the H2A plus. Therefore, at the half equivalence point, we have half a mole of H2A plus and half a mole of HA, which means the concentration of H2A plus is equal to the concentration of HA. And because at this point we have significant amounts of both the weak acid and its conjugate base, we've formed a buffer. So notice how the pH changes very slowly in this region around the half equivalence point one. So this is called buffer region one right in here. And since we've formed a buffer, the Henderson-Hasselbalch equation tells us when the concentration of the weak acid is equal to the concentration of the conjugate base, the pH of the solution is equal to the pKa of the weak acid. In this case, we're talking about the pKa value of the more acidic proton, so the proton that's on the oxygen, NH2A plus. So if we find the half equivalence point and we go over to where this intersects with the y-axis, the pH at this point should be equal to the pKa value for the more acidic proton. So looking at the y-axis here, it looks like it's a little bit over two. And in reality, the value of pKa one is equal to 2.34. Next, let's go back to the first equivalence point, which occurred after we added one mole of hydroxide anions. So at this point, we have HA. If we think about adding another mole of hydroxide anions, so going from one mole total to two moles total, the extra mole of hydroxide anions will completely neutralize the HA and turn it into A minus. So one mole of HA reacts with one mole of hydroxide anions to form one mole of A minus. And since the extra mole of hydroxide anions completely neutralize the HA and turn it into A minus, after we've added a total of two moles of hydroxide anions, we go up to where that intersects our titration curve, and this point represents equivalence point two. So the equivalence point one occurs after adding one mole of hydroxide anions, and we can see the pH by going over to the y-axis. And equivalence point two occurs after we've added a total of two moles of hydroxide anions. Let's go back to our first equivalence point where we had one mole of HA. And at that point, we've already added one mole of hydroxide anions. And so if we add another 0.5 moles of hydroxide anions to give a total of 1.5 moles, we would neutralize half of the HA and turn half of it into A minus. Therefore, we have another half equivalence point after 1.5 moles of hydroxide anions have been added. Let's look in more detail at half equivalence point two, which is right here on our titration curve after we've added 1.5 moles of hydroxide anions. Since we've neutralized half of the HA that was present at equivalence point one, now the concentration of HA is equal to the concentration of A minus. And since we have significant amounts of both the weak acid and its conjugate base, we have a buffer present in the region around half equivalence point two. So we can see the pH is changing very slowly in this region around the half equivalence point as hydroxide anions are added. And so this represents buffer region two. And because the concentration of weak acid is equal to the concentration of its conjugate base, the Henderson-Hasselbalch equation tells us the pH at this point is equal to the pKa value. In this case, it would be pKa two, so one of the acidic protons on the nitrogen. So we can find the value for pKa two by locating half equivalence point two and going over to see where that intersects with our y-axis. It looks to be a little bit under 10, which matches with the actual pKa two value, which turns out to be 9.87. Finally, let's go back to the two equivalence points for our titration curve. The number of equivalence points in a titration curve for a polyprotic acid is equal to the number of acidic protons in the acid. Therefore, since we titrated a diprotic acid with two acidic protons, the titration curve has two equivalence points.