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Continuing our discussion of
acid-base theory,

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I have just drawn here on the
board for you the structure of a

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popular indicator molecule.
Indicator molecules are used

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for titrations.
And titrations will be of focus

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of today's lecture.
This is the indicator molecule,

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phenolphthalein.
It is easier to draw this

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molecular structure than it is
to spell phenolphthalein,

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but you should know how to do
both of these things,

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or at least recognize them.
Now, how do indicator molecules

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work?
Well, during a titration,

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you are mixing chemicals in a
way that leads to a changing pH

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throughout the experiment.
And an indicator molecule like

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phenolphthalein is,
in fact, a Bronsted acid.

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And, if you inspect the
structure that I have drawn here

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for phenolphthalein,
you should be able to begin to

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understand just how it is that a
molecule like this can serve as

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a Bronsted acid.
If I write out the formula,

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you will see that we have quite
a number of carbons and

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hydrogens in the molecule.
And we have oxygens as well.

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Those are the three elements
present in a phenolphthalein

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molecule.
And two of the hydrogens in

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this molecule possess Bronsted
acidity that makes them valuable

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for the function of this system
as an indicator molecule.

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And what happens is when you
have this in solution at low pH,

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it is in this form.
It is neutral.

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And then, as you begin to raise
the pH, at some point a base

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that has a lone pair of
electrons comes along and takes

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one of these two protons in a
molecule, either this one or

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this one.
Those two protons are

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equivalent.
And what I am now going to draw

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is a sequence of arrows
representing the electronic flow

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00:02:52 --> 00:02:57
in the phenolphthalein molecule
that occurs when H plus

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is transferred over to the base
with its lone pair of electrons.

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So, I am starting these arrows
at pairs of electrons and I am

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showing the flow of how they
move.

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00:03:11 --> 00:03:15
And this is a pretty
interesting rearrangement

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because it is complicated.
It is an electronic

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00:03:19 --> 00:03:23
rearrangement that is
transmitted through this

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molecule such that here in the
center, this four coordinate or

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sp three hybridized
carbon, which is bonded to this

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oxygen, has a pair of electrons
that migrates to it.

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And, if I draw now the product
of this reaction,

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what you should be able to see
quite clearly --

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And I am not always explicitly
drawing out the lone pairs of

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electrons on each of the
electronegative oxygen atoms in

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the molecule,
but the sequence of double

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bonds and single bonds in the
molecule has been altered in a

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profound manner because of the
base coming along and removing a

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proton from one of these
peripheral oxygens,

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namely, that one here.
So, we have now a singly

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negatively charged system.
And that charge is balanced,

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over here, by the base that has
the proton attached to it.

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That is the structure-function
relationship that is typical of

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indicator molecules.
And indicator molecules can

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either be naturally occurring,
as they are in some plants.

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Or, alternatively,
they can be molecules that are

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synthesized to respond to
factors like pH in a given way

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that might be desired.
And, in the case of

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phenolphthalein,
this change takes place between

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8.2 to 10.0 pH units.
And the color change associated

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with this is colorless to pink.

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Looking at the structure of the
anionic form of the

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phenolphthalein molecule that
results from its deprotonation,

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what you can see is that this
sp three hybridized

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carbon atom at the center of the
molecule, which in the neutral

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form is acting as an insulator,
preventing the pi-systems of

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the three six-member rings from
communicating,

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is converted into a
three-coordinate sp two

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hybridized carbon atom here at
the center, --

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and provides a conduit for a
communication between the

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pi-systems of the three
six-membered substituted benzene

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rings that are in this molecule.
And that change in electronic

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structure is what leads to the
production of this pink

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Chromaphor in the anionic
deprotonated form of a

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phenolphthalein molecule that is
important at high pH.

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All you have to do,
in fact, is to change the

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substituents on this molecule in
order to get a response at a

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different pH value.
Bromothymol blue is an

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indicator with a structure
similar to that of

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phenolphthalein,
but with a sulfur here in place

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00:07:04 --> 00:07:08
of this carbon.
And in bromothymol blue,

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we also have some isopropyl and
bromine substituents on the

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aromatic rings.
And that modifies its

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properties such that it goes
from yellow to blue as the pH

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rises above 6.0 to about a value
of 7.6.

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00:07:24 --> 00:07:29
So, if you go to a textbook and
look in a table of indicator

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molecules, you will be able to
select an indicator that is

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appropriate for a particular
type of titration.

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00:07:39 --> 00:07:44
And just to give you a very
simple schematic of a classic

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type of titration,
I am going to show you here --

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-- the type of system that you
may need to think about in

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connection with this next
problem set.

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00:08:06 --> 00:08:13
And it is a system like this,
where we have a flask down here

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at the bottom that contains,
initially, some volume of a

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weak acid.

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And that weak acid might be,
for example,

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00:08:30 --> 00:08:34
acetic acid,
as we talked about last time.

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00:08:34 --> 00:08:36
And we will say,
for example,

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that this could be 0.1 molar CH
three COOH.

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That is acetic acid.
And the calculation that we

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left off with at the end of last
hour is one that we will get to

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solving today that had to do
with what is the pH of a tenth

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00:08:56 --> 00:09:01
molar acetic acid solution in
water?

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00:09:01 --> 00:09:04
Or, in other words,
what is the pH of this solution

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00:09:04 --> 00:09:09
down here containing the acetic
acid at the beginning of this

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titration that we are going to
carry out.

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We might have a weak acid down
here.

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We might also have our
indicator present,

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00:09:17 --> 00:09:22
so that we will see a color
change when we pass through a

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particular pH.
And we are going to make the pH

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00:09:27 --> 00:09:30
rise by putting into this
burette here,

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which will allow us to control
a dropwise addition of a strong

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base solution that we are going
to add in and slowly neutralize

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00:09:42 --> 00:09:47
our acetic acid that is down
here at the bottom.

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So, this is strong base.
And, for example,

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that might be tenth molar
sodium hydroxide.

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00:09:55 --> 00:10:00
So, there is a typical
titration.

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00:10:00 --> 00:10:04
And, if you think about it,
because we are adding one

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00:10:04 --> 00:10:07
solution to another,
we are going to have a

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00:10:07 --> 00:10:12
constantly changing volume in
this bottom solution throughout

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00:10:12 --> 00:10:16
the experiment.
And so ultimately what we are

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00:10:16 --> 00:10:21
going to be interested in,
in our questions such as the

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following, as the volume
increases --

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And we will give the volume,
for the purposes of today's

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lecture, the symbol the lower
case letter m -- As the volume

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00:10:42 --> 00:10:44
increases, what happens to the
pH?

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00:10:44 --> 00:10:47
We know that we are using a
strong base to neutralize an

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00:10:47 --> 00:10:51
acid, so the pH is going to
start out low and somehow is

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00:10:51 --> 00:10:53
going to rise.
But what if we wanted to

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predict the mathematical form of
the rise in pH as a function of

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00:10:56 --> 00:11:00
this increasing volume of the
solution?

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00:11:00 --> 00:11:04
Then we would have to have some
equations that would be useful

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for describing this physical
property as a function of this

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00:11:08 --> 00:11:11
increasing volume.
This is what we often want to

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be able to do in chemistry.
We want to be able to describe

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properties that change either as
a function of time or as a

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00:11:18 --> 00:11:21
function of some other variable,
here volume,

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00:11:21 --> 00:11:24
in a titration.
And we want to be able to see

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if we can come up with a set of
equations that would predict

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00:11:28 --> 00:11:34
that change as a function of
this variable that is changing.

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00:11:34 --> 00:11:38
And, in order to do that,
here, I am going to first point

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out that we can put together
some titration equations.

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00:11:43 --> 00:11:56


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00:11:56 --> 00:11:59
And, ultimately,
we are going to want to

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generate a mathematical model
for this titration.

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00:12:05 --> 00:12:07
And then, if you are
experimentalist,

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you want to go ahead and take
that mathematical model and

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compare it to actual
experimental data.

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00:12:15 --> 00:12:18
But let's work on getting a
model generated,

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here.
First we are going to have a

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set of chemical equations.
These correspond to the acid

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HA, which is acetic acid in our
example, in water reacting to

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give some concentration of H
three O plus plus the

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anion A minus,
which is acetate ion,

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in this particular case.
We have a strong base,

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which is sodium hydroxide.
And when you put sodium

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00:13:01 --> 00:13:05
hydroxide in water,
because it is a strong base,

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it ionizes completely.
And I am signifying that with a

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single forward arrow rather than
with an equilibrium arrow.

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This going into water is going
to be Na plus and

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hydroxide, OH minus.
Now, at certain points in a

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titration curve,
you would not need to

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necessarily go much further than
this in order to have enough

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information to start solving for
the pH at a particular value of

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a volume m.
But at some points along that

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titration curve,
that is not enough information.

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And that is because water
itself can sometimes get into

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the equation.
Here is an interesting

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equation, which is H two O plus
H two O going to H three O plus

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00:14:01 --> 00:14:07
plus O minus.

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00:14:07 --> 00:14:12
That is a conceivable reaction
that could be occurring even in

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pure water, but the value of the
equilibrium constant for that

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reaction is very small.
And I will draw that as an

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equilibrium to distinguish it
from the irreversible ionization

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of sodium hydroxide in aqueous
solution.

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I have a pair of equilibria
that I have written.

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I have one forward reaction.
And I want to point out,

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too, that this third equation I
have written here has a special

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name, which is autoprotolysis.

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I will tell you more about the
equilibrium constant associated

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with autoprotolysis in a moment.
But because this equilibrium

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constant is very small,
meaning that there is very

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little H three O plus
and OH minus present in

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pure water, because of that pure
water itself is not a very good

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electrolyte.
Meaning pure water itself is

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not a very good conductor of
electricity, because the

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autoprotolysis reaction lies
mostly over here to the left in

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pure water.
But at certain points along a

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titration curve,
such as the one we are going to

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00:15:26 --> 00:15:31
want to develop on this panel
here, we are going to need to

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include autoprotolysis in order
to get the right answer because

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it is there, it can occur.
Let's continue.

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00:16:02 --> 00:16:08
Another type of equation that
we can use to put this into some

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00:16:08 --> 00:16:14
kind of a mathematical footing
is the balance of charge in a

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00:16:14 --> 00:16:20
solution like the one that we
are talking about in that beaker

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00:16:20 --> 00:16:25
to which we are adding sodium
hydroxide solution.

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00:16:25 --> 00:16:31
And the charge balance
consideration tells us that the

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00:16:31 --> 00:16:37
concentration of hydronium ion
plus the concentration of sodium

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00:16:37 --> 00:16:43
ion must be equal to the
concentration of hydroxide plus

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00:16:43 --> 00:16:50
the concentration of acetate
ion, A minus.

213
00:16:50 --> 00:16:53
Because what I have done here
in the charge balance

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00:16:53 --> 00:16:57
consideration is I have said
okay, what are all the possible

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00:16:57 --> 00:17:01
charged species in solution?
I can make a list of them.

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00:17:01 --> 00:17:05
It is those four things.
And I know that the number of

217
00:17:05 --> 00:17:08
positive charges must equal the
number of negative charges

218
00:17:08 --> 00:17:12
because this is overall a
neutral beaker in which we are

219
00:17:12 --> 00:17:15
putting things.
But everything that you put in

220
00:17:15 --> 00:17:18
is charge neutral,
so the positives have to equal

221
00:17:18 --> 00:17:22
the negatives.
That is a very nice limiting

222
00:17:22 --> 00:17:27
equation that helps us
understand how the different

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competing chemical reactions and
equilibria will all settle down

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and arrive at the physical
result, which will be our

225
00:17:39 --> 00:17:43
observable.
And then, in addition to that,

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00:17:43 --> 00:17:46
we have a mass balance
consideration.

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00:17:46 --> 00:17:50
And in this mass balance
consideration,

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00:17:50 --> 00:17:57
we will be able to write that
our HA initial value --

229
00:17:57 --> 00:18:01
-- that is square brackets,
once again, denoting

230
00:18:01 --> 00:18:06
concentration -- will be,
at any point in time,

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00:18:06 --> 00:18:11
equal to the actual value of HA
plus the concentration of A

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00:18:11 --> 00:18:15
minus.
And that just says that when

233
00:18:15 --> 00:18:20
you have acetic acid and you put
it into water,

234
00:18:20 --> 00:18:25
it ionizes partly to hydronium
ion and acetate ion.

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00:18:25 --> 00:18:31
And that which does not ionize
is unionized HA.

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00:18:31 --> 00:18:36
And so, the ionized A minus
concentration plus the

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00:18:36 --> 00:18:43
unionized HA concentration is
equal to the sum total of the

238
00:18:43 --> 00:18:47
acetic acid that is present in
solution.

239
00:18:47 --> 00:18:51
And, similarly,
we have another mass balance

240
00:18:51 --> 00:18:57
equation, which is just that the
sodium hydroxide initial is

241
00:18:57 --> 00:19:03
equal to our sodium ion
concentration because all of it

242
00:19:03 --> 00:19:09
is ionized.
That is a little simpler than

243
00:19:09 --> 00:19:13
the situation with the acetic
acid.

244
00:19:13 --> 00:19:20
And then, in addition to those
equations, we have our

245
00:19:20 --> 00:19:24
equilibrium equations.

246
00:19:24 --> 00:19:37


247
00:19:37 --> 00:19:42
And we are going to have two of
these because there are two

248
00:19:42 --> 00:19:47
equilibria that I wrote over
there under titration equations.

249
00:19:47 --> 00:19:53
The first of these is going to
have the equilibrium constant

250
00:19:53 --> 00:19:58
Ka, and that is going to be H
three O plus times A

251
00:19:58 --> 00:20:02
minus,
which is acetate in the case of

252
00:20:02 --> 00:20:09
acetic acid, all over HA.
That is the expression for the

253
00:20:09 --> 00:20:14
equilibrium of acetic acid
ionization in dilute aqueous

254
00:20:14 --> 00:20:18
solution.
And I will give you a value for

255
00:20:18 --> 00:20:23
this in a moment.
I won't write it right here.

256
00:20:23 --> 00:20:28
Well, why don't I just do that.
It is 1.8x10^-5 for this

257
00:20:28 --> 00:20:35
specific example of acetic acid.
And then, if you were working

258
00:20:35 --> 00:20:39
with some other weak acid,
you would be able to go to a

259
00:20:39 --> 00:20:44
table and look up the Ka value
for it, if it has been measured

260
00:20:44 --> 00:20:48
and reported and tabulated.
And then, we have Kw.

261
00:20:48 --> 00:20:52
And this is the equilibrium
expression for the

262
00:20:52 --> 00:20:56
autoprotolysis reaction that I
mentioned over there,

263
00:20:56 --> 00:21:00
the other equilibrium,
which is H three O plus times

264
00:21:00 --> 00:21:06
OH minus.
And that is equal to 1x10^-14,

265
00:21:06 --> 00:21:09
so a very small equilibrium
constant.

266
00:21:09 --> 00:21:13
And some of you may be
wondering why water is not

267
00:21:13 --> 00:21:17
appearing in an expression like
this one or like this one.

268
00:21:17 --> 00:21:22
And the reason for this is that
in the expressions that I am

269
00:21:22 --> 00:21:26
deriving here,
we are using concentrations in

270
00:21:26 --> 00:21:30
place of activities for these
species.

271
00:21:30 --> 00:21:35
In the derivation of the
equilibrium constants,

272
00:21:35 --> 00:21:40
the activity of pure water at
very high concentration is

273
00:21:40 --> 00:21:46
approximated as one.
And so, this is simplified as

274
00:21:46 --> 00:21:50
shown here.
And having put all these

275
00:21:50 --> 00:21:55
equations together,
what we are going to want to do

276
00:21:55 --> 00:22:01
is take the following approach.
We are looking at the

277
00:22:01 --> 00:22:04
expression for
electro-neutrality here,

278
00:22:04 --> 00:22:08
the equal number of positively
and negatively charged species

279
00:22:08 --> 00:22:11
in solution.
And I want to get this

280
00:22:11 --> 00:22:15
expression rewritten in terms of
things that I know at any point

281
00:22:15 --> 00:22:20
along the titration curve,
with the exception of the one

282
00:22:20 --> 00:22:22
thing that I want to know,
the pH.

283
00:22:22 --> 00:22:26
And the pH we can calculate
easily if we know the H three O

284
00:22:26 --> 00:22:32
plus concentration.
So, I am going to be seeking to

285
00:22:32 --> 00:22:35
get Na plus,
OH minus,

286
00:22:35 --> 00:22:39
and A minus rewritten
in terms of H three O plus

287
00:22:39 --> 00:22:44
and constants using
the equations that I have up

288
00:22:44 --> 00:22:46
here on the board.

289
00:22:46 --> 00:23:00


290
00:23:00 --> 00:23:05
First, let's attempt to do that
for A minus using this

291
00:23:05 --> 00:23:10
Ka expression.
And I am going to rearrange it.

292
00:23:10 --> 00:23:19


293
00:23:19 --> 00:23:27
And it is going to be equal to
Ka times HA divided by H three O

294
00:23:27 --> 00:23:32
plus.
But I still have a problem with

295
00:23:32 --> 00:23:37
this expression for A minus
because I am going to

296
00:23:37 --> 00:23:42
need to replace HA with
something that will allow me to

297
00:23:42 --> 00:23:47
get everything in terms of just
H three O plus and

298
00:23:47 --> 00:23:50
constants.
And so, if I look over here at

299
00:23:50 --> 00:23:55
the mass balance equation,
I can use this one and replace

300
00:23:55 --> 00:24:01
HA with HA initial minus
A minus.

301
00:24:01 --> 00:24:44


302
00:24:44 --> 00:24:47
Now I have A minus
written in terms of an initial

303
00:24:47 --> 00:24:49
concentration of HA itself and
constants.

304
00:24:49 --> 00:24:52
I need to rearrange this a
little bit.

305
00:24:52 --> 00:25:00
I am going to multiply through
by H three O plus so

306
00:25:00 --> 00:25:08
that I can write A minus times H
three O plus is equal to Ka

307
00:25:08 --> 00:25:16
times HA initial minus Ka times
A minus.

308
00:25:16 --> 00:25:24
And hopefully I have done that

309
00:25:24 --> 00:25:30
right.
And, if so, I will be able to

310
00:25:30 --> 00:25:39
say that A minus is going to be
equal to Ka times HA initial all

311
00:25:39 --> 00:25:45
over H three O plus.

312
00:25:45 --> 00:25:48


313
00:25:48 --> 00:25:58


314
00:25:58 --> 00:26:03
Does that look good?
Plus Ka, thank you.

315
00:26:03 --> 00:26:08
H3O+ plus Ka.

316
00:26:08 --> 00:26:15
And let's bring
that down here.

317
00:26:15 --> 00:26:25


318
00:26:25 --> 00:26:28
Because now,
I would like to go ahead and

319
00:26:28 --> 00:26:32
make this substitution for the
charge.

320
00:26:32 --> 00:26:38
We have our A minus.
The other ones are a little

321
00:26:38 --> 00:26:41
easier.
We now have this equation H

322
00:26:41 --> 00:26:48
three O plus.
And we were going to replace Na

323
00:26:48 --> 00:26:53
plus by NaOH initial.

324
00:26:53 --> 00:27:00
And we are going to recognize
that we can use the equation for

325
00:27:00 --> 00:27:06
autoprotolysis to replace
hydroxide with Kw divided by H

326
00:27:06 --> 00:27:14
three O plus.
And then now we have the A

327
00:27:14 --> 00:27:21
minus being replaced
by Ka HA initial all over H

328
00:27:21 --> 00:27:29
three O plus plus Ka.

329
00:27:29 --> 00:27:34
And at this point,
you have actually done a really

330
00:27:34 --> 00:27:39
nice thing, because you have
generated a general titration

331
00:27:39 --> 00:27:45
equation for the titration of
interest in this particular

332
00:27:45 --> 00:27:49
case.
You can see that at any point

333
00:27:49 --> 00:27:54
along that curve for a given
volume m, we can now stop and

334
00:27:54 --> 00:28:00
walk along in one milliliter
increments.

335
00:28:00 --> 00:28:03
And, in each spot along that
curve, we could go ahead and

336
00:28:03 --> 00:28:07
solve this equation to get the
value of H three O plus.

337
00:28:07 --> 00:28:09
And then we would just have to

338
00:28:09 --> 00:28:13
take the negative log of that,
and we would have the pH at

339
00:28:13 --> 00:28:16
that position.
And we are interested if this

340
00:28:16 --> 00:28:19
type of a model for the
mathematical form of the

341
00:28:19 --> 00:28:23
titration curve bears any
resemblance to what one sees in

342
00:28:23 --> 00:28:27
reality.
And, if so, one could then be,

343
00:28:27 --> 00:28:31
at least for the time being,
until your theory no longer

344
00:28:31 --> 00:28:36
fits, satisfied that you have
accounted for the various

345
00:28:36 --> 00:28:41
equilibria that could be present
in a system such as the one we

346
00:28:41 --> 00:28:44
are describing here.
This happens to be a cubic

347
00:28:44 --> 00:28:48
equation, and it has,
therefore, three roots.

348
00:28:48 --> 00:28:53
And when we solve it at each of
these points along the titration

349
00:28:53 --> 00:28:57
equation, because of the
physically realistic quantities

350
00:28:57 --> 00:29:00
that we are going to be
interested in,

351
00:29:00 --> 00:29:05
only one of the roots is
positive.

352
00:29:05 --> 00:29:09
We are talking about H three O
plus concentration.

353
00:29:09 --> 00:29:11
It cannot be a negative
concentration.

354
00:29:11 --> 00:29:15
The positive root is the one we
want, and then we can convert

355
00:29:15 --> 00:29:19
that to pH and can see what kind
of result we get.

356
00:29:19 --> 00:29:22
And so, I will show you how we
can do that using a tool

357
00:29:22 --> 00:29:25
available on Athena.

358
00:29:25 --> 00:30:03


359
00:30:03 --> 00:30:07
First of all,
let me tell you that for the

360
00:30:07 --> 00:30:13
purposes of doing this,
I am going to replace these

361
00:30:13 --> 00:30:17
various quantities by some
simpler symbols,

362
00:30:17 --> 00:30:21
here.
I am going to go x plus a,

363
00:30:21 --> 00:30:27
a is our NaOH initial,
is equal to (b over x) plus (c

364
00:30:27 --> 00:30:35
times d) over (x plus d).

365
00:30:35 --> 00:30:38
So, I am going to use those
simpler symbols for this

366
00:30:38 --> 00:30:40
equation.
I could rearrange it and show

367
00:30:40 --> 00:30:43
you how it is a cubic equation,
you set it equal to zero,

368
00:30:43 --> 00:30:46
and then you solve,
and you get the roots.

369
00:30:46 --> 00:30:49
Just like a quadratic equation
but with another term.

370
00:30:49 --> 00:30:52
And, instead of doing that,
we are just going to leave it

371
00:30:52 --> 00:30:54
in this form,
which is simpler to look at.

372
00:30:54 --> 00:30:56
And we are just going to use
that.

373
00:30:56 --> 00:31:00
And so, let's go ahead and
define A.

374
00:31:00 --> 00:31:03
That is the initial
concentration of sodium

375
00:31:03 --> 00:31:06
hydroxide.
We are going to go 0.1.

376
00:31:06 --> 00:31:12
And remember I need to get this
in terms of the quantity m here,

377
00:31:12 --> 00:31:16
which is the volume.
And so, I am going to take m,

378
00:31:16 --> 00:31:20
minus 20 over m.
And the reason I am doing is

379
00:31:20 --> 00:31:26
this is I am saying that we have
an 0.1 molar solution of sodium

380
00:31:26 --> 00:31:32
hydroxide that we are adding.
But there is a dilution factor

381
00:31:32 --> 00:31:36
that you have to take care of.
So, at every value m we are

382
00:31:36 --> 00:31:41
saying that we are starting with
a volume of 20 in the acetic

383
00:31:41 --> 00:31:44
acid flask at the bottom there
at the beginning.

384
00:31:44 --> 00:31:48
And then we are starting to add
volume units of sodium

385
00:31:48 --> 00:31:50
hydroxide.
And so the volume,

386
00:31:50 --> 00:31:53
m, is increasing.
That is how I take care of

387
00:31:53 --> 00:31:57
that, there.
And then our value of b is Kw,

388
00:31:57 --> 00:32:01
so that is 1 E minus 14.

389
00:32:01 --> 00:32:05
That is for autoprotolysis.
And then we have c,

390
00:32:05 --> 00:32:09
that is our initial
concentration of acetic acid.

391
00:32:09 --> 00:32:14
And we happen to have a
dilution factor here,

392
00:32:14 --> 00:32:18
too, but it is simpler,
just 20 divided by m.

393
00:32:18 --> 00:32:23
And there is that.
And then d, what did I say that

394
00:32:23 --> 00:32:26
d was?
d is our equilibrium constant

395
00:32:26 --> 00:32:31
for HA, 1.8 E minus 5.

396
00:32:31 --> 00:32:35
There is our acidity constant
for acetic acids.

397
00:32:35 --> 00:32:40
Now I have the four things in
there that I need,

398
00:32:40 --> 00:32:46
and those are four things that
I know because of the initial

399
00:32:46 --> 00:32:51
conditions that I am defining
for this problem.

400
00:32:51 --> 00:32:57
And then I am going to define
my equation as being equal to x

401
00:32:57 --> 00:33:03
plus a is equal to (b over x)
plus (c times d) all over (x

402
00:33:03 --> 00:33:07
plus d).
There is my equation.

403
00:33:07 --> 00:33:11
And this nice calculator,
all I have to do,

404
00:33:11 --> 00:33:14
in principle,
is tell it what the value of m

405
00:33:14 --> 00:33:17
is.
And then it is going to solve

406
00:33:17 --> 00:33:21
that cubic equation,
give me back the three roots.

407
00:33:21 --> 00:33:25
I take the positive one and I
take the negative log of it,

408
00:33:25 --> 00:33:30
and then I know the pH at that
value of m.

409
00:33:30 --> 00:33:34
I can develop a whole titration
curve based on the equations

410
00:33:34 --> 00:33:38
that we have been discussing.
And so let's choose a value of

411
00:33:38 --> 00:33:41
m here to start with,
and I am going to choose 30.

412
00:33:41 --> 00:33:45
We are starting at 20.
At the very beginning of the

413
00:33:45 --> 00:33:48
titration, we are at 20.
Let's see what happens at 30

414
00:33:48 --> 00:33:52
given the particular value of m.
Here is what I do.

415
00:33:52 --> 00:33:54
There may be other ways to do
this.

416
00:33:54 --> 00:33:59
I am defining this thing called
the solution set.

417
00:33:59 --> 00:34:03
Because the solving of that
equation is going to give us

418
00:34:03 --> 00:34:07
three values.
And then I want to be able to

419
00:34:07 --> 00:34:12
pick the one that I want.
We would use sol-set is equal

420
00:34:12 --> 00:34:16
to, we are going to solve that
equation, and we want x,

421
00:34:16 --> 00:34:21
which is our H three O plus
concentration.

422
00:34:21 --> 00:34:25
And sometimes it chokes.
Let's see what happens.

423
00:34:25 --> 00:34:31
Sometimes it is a little slow.
But it takes a lot longer if

424
00:34:31 --> 00:34:35
you actually try to work it all
out by hand.

425
00:34:35 --> 00:34:40


426
00:34:40 --> 00:34:44
I have a picture of an image of
this that I grabbed.

427
00:34:44 --> 00:34:46
I can show you if this fails on
me.

428
00:34:46 --> 00:34:49
Don't fail on me.
There we go.

429
00:34:49 --> 00:34:54
You can see it got the answer.
And so here is a positive root,

430
00:34:54 --> 00:34:59
and then there are two negative
roots to this cubic equation in

431
00:34:59 --> 00:35:03
x.
And I want to find out the pH.

432
00:35:03 --> 00:35:07
When we are up to 30 volume
units, m equals 30.

433
00:35:07 --> 00:35:13
And so, I will define p here as
equal to the negative log base

434
00:35:13 --> 00:35:18
10 of the first of the answers
in my solution set,

435
00:35:18 --> 00:35:23
sol-set one,
which is the positive x value.

436
00:35:23 --> 00:35:27
And, bingo, there is the pH,
4.745 whatever dot,

437
00:35:27 --> 00:35:32
dot, dot.
That is now the pH at a volume

438
00:35:32 --> 00:35:36
of 30.
Now, you figure I am starting

439
00:35:36 --> 00:35:40
with 20 volume units of 0.1
molar acetic acid.

440
00:35:40 --> 00:35:43
And by the time m equals 30
volume units,

441
00:35:43 --> 00:35:49
that means I have added ten
volume units of strong base,

442
00:35:49 --> 00:35:52
enough to quench half of my
acetic acid.

443
00:35:52 --> 00:35:57
And what if I had just only
quenched none of it?

444
00:35:57 --> 00:36:02
I can change m now.
Once you have this in here,

445
00:36:02 --> 00:36:04
you don't have to retype these
things.

446
00:36:04 --> 00:36:07
I can go back here and say let
m be 20.

447
00:36:07 --> 00:36:11
And, if I do that,
now I am actually at the zero

448
00:36:11 --> 00:36:13
point, before I add any strong
base.

449
00:36:13.831 --> 2.87.
And there is the pH now,

450
2.87. --> 00:36:16


451
00:36:16 --> 00:36:20
The answer to the expression
that we had at the very end of

452
00:36:20 --> 00:36:24
class last time that was going
to tell us what is the pH of

453
00:36:24 --> 00:36:28
tenth molar acetic acid,
there is the answer,

454
00:36:28 --> 00:36:33
2.87, because that is what we
have when we start.

455
00:36:33 --> 00:36:40
Now you can see that you could
just go through and use that to

456
00:36:40 --> 00:36:46
get the pH at any value of m
along this titration curve.

457
00:36:46 --> 00:36:52
And when you do that,
let's show you what the result

458
00:36:52 --> 00:36:57
is, now what you see here is a
set of data.

459
00:36:57 --> 00:37:02
I start at 20.
And I'm going up in single

460
00:37:02 --> 00:37:04
volume units,
21, 22, 23, and 24,

461
00:37:04 --> 00:37:09
I am showing what the pH is at
each of those choices of m.

462
00:37:09 --> 00:37:13
And I have gone all the way
from 20 to 60.

463
00:37:13 --> 00:37:16
And you take that data set --

464
00:37:16 --> 00:37:42


465
00:37:42 --> 00:37:51
And then you can take a file,
like the one I just showed you,

466
00:37:51 --> 00:38:00
which just had values of m and
corresponding values of pH.

467
00:38:00 --> 00:38:02
And you can go ahead and just
plot that thing.

468
00:38:02 --> 00:38:05
And, when you do that,
you get the form of the

469
00:38:05 --> 00:38:09
titration curve here,
which is the classic form of a

470
00:38:09 --> 00:38:12
titration curve for titrating a
weak acid with a strong base.

471
00:38:12 --> 00:38:15
And you see that here,
m was 20, and we let it

472
00:38:15 --> 00:38:18
increment through all the way up
to 60.

473
00:38:18 --> 00:38:21
And what you see initially,
and you are going to need to

474
00:38:21 --> 00:38:25
study this one and the other
cases of titration curves that

475
00:38:25 --> 00:38:29
are in that chapter of your
book, is how the little subtle

476
00:38:29 --> 00:38:33
features differ from one case to
another.

477
00:38:33 --> 00:38:36
They do differ.
A much simpler titration curve

478
00:38:36 --> 00:38:40
than this one is obtained if you
titrate a strong acid with a

479
00:38:40 --> 00:38:43
strong base.
This is titration of a weak

480
00:38:43 --> 00:38:46
acid with a strong base.
And you see that initially,

481
00:38:46 --> 00:38:50
we have that 2.87 initial pH
value, that it kind of shoots up

482
00:38:50 --> 00:38:54
quickly here at the beginning.
There is a steep rise,

483
00:38:54 --> 00:38:57
initially.
And then it kind of levels off.

484
00:38:57 --> 00:39:01
And in this region here,
the pH is not changing very

485
00:39:01 --> 00:39:06
rapidly at all as you are adding
more sodium hydroxide.

486
00:39:06 --> 00:39:09
Until you get out here,
and the pH just shoots up.

487
00:39:09 --> 00:39:13
And you will notice that it is
around where we get to 40.

488
00:39:13 --> 00:39:18
Because we started with 20 mLs
of tenth molar acetic acid.

489
00:39:18 --> 00:39:22
And, when we get out to m
equals 40, we have added 20 mLs

490
00:39:22 --> 00:39:26
of tenth molar sodium hydroxide.
At that point the pH just

491
00:39:26 --> 00:39:29
shoots up.
And then over here it turns

492
00:39:29 --> 00:39:33
over and just goes kind of
slowly for a minute.

493
00:39:33 --> 00:39:37
And it just approaches an
asymptotic limiting value over

494
00:39:37 --> 00:39:39
here on the high pH side of the
scale.

495
00:39:39 --> 00:39:44
One thing you will notice here
is that at this point where we

496
00:39:44 --> 00:39:48
have added the same number of
equivalence of sodium hydroxide

497
00:39:48 --> 00:39:52
molecules as we had acetic acid
molecules to start with,

498
00:39:52 --> 00:39:56
this occurs over here at
greater than 8.0 pH units rather

499
00:39:56 --> 7.0.
than at the neutral value of

500
7.0. --> 00:40:00


501
00:40:00 --> 00:40:03
And I would like you to think
about why that might be.

502
00:40:03 --> 00:40:08
And that will be something that
will be covered in recitation,

503
00:40:08 --> 00:40:12
but it is important.
If we had derived here today

504
00:40:12 --> 00:40:16
the expression for the titration
of a strong acid with a strong

505
00:40:16 --> 00:40:19
base, then at 40 we would be
right at 7.0,

506
00:40:19 --> 00:40:23
but when we were using this
weak acid, acetic acid,

507
00:40:23 --> 00:40:27
it comes up here over 8.0 pH
units.

508
00:40:27 --> 00:40:31
This point where we have added
as many strong base molecules as

509
00:40:31 --> 00:40:33
we started with,
weak acid molecules,

510
00:40:33 --> 00:40:38
is called the equivalence or
the stoichiometric point in the

511
00:40:38 --> 00:40:40
titration.
Moreover, where would your

512
00:40:40 --> 00:40:43
indicator molecule change color?
At what point,

513
00:40:43 --> 00:40:45
at what value of m,
for example?

514
00:40:45 --> 00:40:49
It would be close to there.
It would be because,

515
00:40:49 --> 00:40:53
in the case of phenolphthalein,
it changes between right in

516
00:40:53 --> 00:40:55
here.
It might change right around

517
00:40:55 --> 00:41:00
maybe 41 or so.
m equals 41 might be where the

518
00:41:00 --> 00:41:04
color change from colorless to
pink occurs, for the reasons

519
00:41:04 --> 00:41:07
that we have discussed.
And then right here,

520
00:41:07 --> 00:41:11
it turns out at this value of m
equals 30, the pH is equal to

521
00:41:11 --> 00:41:14
the pKa of acetic acid.
And that is a relationship that

522
00:41:14 --> 00:41:19
I would like you to think about.
And it is a relationship that

523
00:41:19 --> 00:41:22
comes into play very strongly
when we talk about the

524
00:41:22 --> 00:41:25
preparation of buffer solutions
according to the

525
00:41:25 --> 00:41:30
Henderson-Hasselbach equation.
At m equals 30,

526
00:41:30 --> 00:41:34
the pH value of the titration
is the pKa of acetic acid,

527
00:41:34 --> 00:41:37
which is right here at about
4.7 or so.

528
00:41:37 --> 00:41:41
We did that one a moment ago.
We found out what that value

529
00:41:41 --> 00:41:43
was.
And, in plus or minus 1.0 pH

530
00:41:43 --> 00:41:46
units of that,
we have the part of the

531
00:41:46 --> 00:41:50
titration curve that is called
the buffering region.

532
00:41:50 --> 00:41:53
In here we have generated a
buffer in situ.

533
00:41:53 --> 00:41:57
And buffers help the pH not to
change too much when acids or

534
00:41:57 --> 00:42:02
bases are added in solution.
That is what buffers do.

535
00:42:02 --> 00:42:06
They try to help you keep in a
constant pH range because lots

536
00:42:06 --> 00:42:10
of things are dependent upon a
particular pH for proper

537
00:42:10 --> 00:42:12
functioning.
And you know that this is

538
00:42:12 --> 00:42:16
important in the world's oceans,
it is important in our

539
00:42:16 --> 00:42:19
environment, and it is important
in our bodies.

540
00:42:19 --> 00:42:23
So, the pH considerations for
aqueous solutions are quite

541
00:42:23 --> 00:42:26
important.
And I hope that today I have

542
00:42:26 --> 00:42:31
been able to give you a nice way
of thinking about some of the

543
00:42:31 --> 00:42:33
math that goes behind a
titration curve.

544
00:42:33.691 --> 42:36
Have a nice Halloween.