Our seventh learning objective is
dealing with polyprotic acids and looking at
titration curves of these polyprotic acids.
The polyprotic means you have more then
one proton that can be donated from this acid.
They call that being a liable proton
donatable proton.
So we are looking at a titration donatable proton.
So we are looking at a titration
curve on the screen right now
of a diprotic acid being
titrated with a strong base.
What we want to be able to do is recognize what titrated with a strong base.
What we want to be able to do is recognize what
species would be present at every point
along the titration curve.
We will consider the pHs at those equivalence point
but we will not actually calculate the pHs
during a polyprotic acid titration.
As we look at this titration curve
we have a diprotic acid.
A diprotic acid we are goign to be looking at
is carbonic acid.
It is going to be reacting sodium hydroxide. is carbonic acid.
It is going to be reacting sodium hydroxide.
That is what is occurring during this titration curve.
So we see that we will have two equivalence points.
We have one here, we have one here
and we have the pH changing
gradual fashions
and then spiking up, and then gradual fashions gradual fashions
and then spiking up, and then gradual fashions
and spiking up during the course of this titration curve. and then spiking up, and then gradual fashions
and spiking up during the course of this titration curve.
Lets begin by looking at the over all reaction of
carbonic acid with sodium hydroxide.
This is the overall reaction. You are going to need two
sodium hydroxides for every
carbonic acid
moles that you have in there.
If we wrote the net ionic equation we
still see that 2 to 1 ratio
between the OH-
and the H_2CO_3. between the OH-
and the H_2CO_3.
But when this reaction takes place it
the NaOH doesn't pull off both
protons off the carbonic gas
at a time. What it does
is occurs in a step-wise fashion.
What to I mean by that? Well, the first step will be
removal of the first proton.
It will be the easiest one to
pull off of the H_2CO_3.
So we will be removing one of these
protons and forming
the bicarbonate ion
during this first titration curve.
So we begin by adding
some base. It will convert some of
the acid over to its conjugate base.
We will have a range where we will have a buffer. the acid over to its conjugate base.
We will have a range where we will have a buffer.
Because there will be both We will have a range where we will have a buffer.
Because there will be both
the H_2CO_3 and the HCO_3-
present in solution
and we will see that buffer occur.
So it will be a gradual increase
and eventually you will be at the equivalence point.
When you are at the equivalence point you will have converted all of your
base and all of you H_2CO_3
over to the bicarbonate ion.
Now this bicarbonate ion
is still able to donate a proton Now this bicarbonate ion
is still able to donate a proton
so we still expect the pH to be in a
acid range, because it will behave as an acid.
So lets look at the curve.
We are looking at the area inside this blue box.
And we are once again considering
H_2CO_3 And we are once again considering
H_2CO_3
and one of the OH-'s
converting over to the bicarbonate ion
and water.
So we see this area of
a flattening out sort of.
That is occurring at the buffer, so we have a
buffer between then That is occurring at the buffer, so we have a
buffer between then
bicarbonate ion and the
carbonic acid here. bicarbonate ion and the
carbonic acid here.
We can use the Henderson-Hasselbalch equation to calculate the pHs
during that range if we were asked to.
But like I said I won't ask you to calculate
pHs of these diprotic acids.
We hit this equivalence point and when we are at the equivalence point we pHs of these diprotic acids.
We hit this equivalence point and when we are at the equivalence point we
see that we still have a pH in a We hit this equivalence point and when we are at the equivalence point we
see that we still have a pH in a
acid range because this thing can
still behave as an acid.
So we get to this first equivalence point and then we still behave as an acid.
So we get to this first equivalence point and then we
continue adding OH- and then at that
point it is going to start taking it off on the second
the second hydrogen. So we have carbonate ion that got converted.
In the first equivalence point
the OH is continually being added. In the first equivalence point
the OH is continually being added.
And we are going to be forming the carbonate ion
as we continue on this reaction. And we are going to be forming the carbonate ion
as we continue on this reaction.
Dr. Allison SoultLecturer
Dr. Kim WoodrumSr. Lecturer
