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Titrations of polyprotic acids
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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Just out of curiosity, when adding the first mol of OH- to the H2A solution, as the H2A is being converted into HA will any of the OH- neutralized the HA into A-? or will the OH- solely neutralize the H2A until it is exhausted then move on to the HA?(1 vote)
The base, OH^(-), will deprotonate the diprotic species, H2A, exclusively before it begins deprotonating the monoprotic species, HA^(-). This is because the first proton of a polyprotic acid is significantly more acidic than the second. The first acidic proton requires a minor amount of base to react so the H2A will react with any added base, and neutralize it, before it has a chance to react with any HA^(-). Only once all the H2A has been converted to HA^(-) and the amount of base is much greater can the second acid-base reaction begin.
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)
Video transcript
- [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.