- [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.