Titrations of polyprotic acids | Acids and bases | AP Chemistry | Khan Academy

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⁺.

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; 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, converting the protonated form of alanine into the overall neutral form of alanine, HA. Let's say that we start out in our titration with 1 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⁺ is converted into HA. So, 1 mole of H₂A⁺ reacts with 1 mole of hydroxide anion to form 1 mole of HA. Now we have 1 mole of HA in solution. 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⁻.

A⁻ is the conjugate base to HA. It would take another mole of hydroxide anions to completely neutralize the HA and turn it into A⁻. So, 1 mole of HA reacts with 1 mole of hydroxide anion to form 1 mole of A⁻. 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. 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 are moles of hydroxide anions added. Before we've added any hydroxide anions, at this point right here on our titration curve, we have our diprotic acid present, so H₂A⁺. Since we started with 1 mole of H₂A⁺, it would take 1 mole of hydroxide anion to completely neutralize the H₂A⁺.

After one mole of hydroxide anions has been added, that brings us to equivalence point 1. So, we go to one mole of hydroxide anions on the x-axis, and we go up to where that intersects our titration curve. This point on our titration curve is equivalence point 1. At equivalence point 1, all the H₂A⁺ has been converted into HA, so now we have 1 mole of HA present.

Next, let's think about adding 0.5 moles of hydroxide anions. That would only neutralize half of the H₂A⁺ that was initially present. Therefore, if we go to 0.5 moles of hydroxide anions and we go up to where that intersects our titration curve, this point represents the half equivalence point, or half equivalence point 1, for this titration.

Next, let's look at half equivalence point 1 in more detail. Here is half equivalence point 1 on the titration curve. If we started out with 1 mole of H₂A⁺ and we've added 0.5 moles of hydroxide anions, we've neutralized half, or 0.5 moles, of the H₂A⁺. Therefore, at the half equivalence point, we have half a mole of H₂A⁺ and half a mole of HA, which means the concentration of H₂A⁺ is equal to the concentration of HA.

Because at this point we have significant amounts of both a weak acid and its conjugate base, we formed a buffer. Notice how the pH changes very slowly in this region around the half equivalence point 1. This is called buffer region 1 right in here. Since we formed a buffer, the Henderson-Hasselbalch equation tells us that 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 in H₂A⁺. 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 2, and in reality, the value of pKa1 is equal to 2.334.

Next, let's go back to the first equivalence point, which occurred after we added 1 mole of hydroxide anion. At this point, we have HA. If we think about adding another mole of hydroxide anion, going from one mole total to 2 moles total, the extra mole of hydroxide anions will completely neutralize the HA and turn it into A⁻.

So, 1 mole of HA reacts with 1 mole of hydroxide anion to form 1 mole of A⁻. Since the extra mole of hydroxide anions completely neutralized the HA and turned it into A⁻, after we've added a total of 2 moles of hydroxide anions, we go up to where that intersects our titration curve, and this point represents equivalence point 2.

The equivalence point 1 occurs after adding 1 mole of hydroxide anions, and we can see the pH by going over to the y-axis. Equivalence point 2 occurs after we've added a total of 2 moles of hydroxide anions. Let's go back to our first equivalence point where we had 1 mole of HA. At that point, we've already added 1 mole of hydroxide anions.

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⁻. 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 2, 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 1, now the concentration of HA is equal to the concentration of A⁻. Since we have significant amounts of both a weak acid and its conjugate base, we have a buffer present in the region around half equivalence point 2.

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 2. 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 pKa2, so one of the acidic protons on the nitrogen.

We can find the value for pKa2 by locating half equivalence point 2 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 pKa2 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.