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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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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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And this is a pretty
interesting rearrangement

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because it is complicated.
It is an electronic

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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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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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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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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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00:08:06,000 --> 00:08:13,000
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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00:08:27,000 --> 00:08:30,000
And that weak acid might be,
for example,

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acetic acid,
as we talked about last time.

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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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molar acetic acid solution in
water?

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00:09:01,000 --> 00:09:04,000
Or, in other words,
what is the pH of this solution

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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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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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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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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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So, there is a typical
titration.

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00:10:00,000 --> 00:10:04,000
And, if you think about it,
because we are adding one

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solution to another,
we are going to have a

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constantly changing volume in
this bottom solution throughout

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the experiment.
And so ultimately what we are

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going to be interested in,
in our questions such as the

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following, as the volume
increases --

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00:10:36,000 --> 00:10:39,000
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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increases, what happens to the
pH?

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We know that we are using a
strong base to neutralize an

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acid, so the pH is going to
start out low and somehow is

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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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00:10:56,000 --> 00:11:00,000
this increasing volume of the
solution?

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00:11:00,000 --> 00:11:04,000
Then we would have to have some
equations that would be useful

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00:11:04,000 --> 00:11:08,000
for describing this physical
property as a function of this

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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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function of some other variable,
here volume,

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00:11:21,000 --> 00:11:24,000
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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00:11:28,000 --> 00:11:34,000
that change as a function of
this variable that is changing.

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00:11:34,000 --> 00:11:38,000
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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00:11:56,000 --> 00:11:59,000
And, ultimately,
we are going to want to

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generate a mathematical model
for this titration.

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00:12:05,000 --> 00:12:07,000
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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00:12:15,000 --> 00:12:18,000
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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00:12:55,000 --> 00:13:01,000
which is sodium hydroxide.
And when you put sodium

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00:13:01,000 --> 00:13:05,000
hydroxide in water,
because it is a strong base,

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00:13:05,000 --> 00:13:09,000
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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00:13:18,000 --> 00:13:22,000
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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00:14:01,000 --> 00:14:07,000
plus O minus.

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00:14:07,000 --> 00:14:12,000
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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00:14:21,000 --> 00:14:26,000
equilibrium to distinguish it
from the irreversible ionization

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of sodium hydroxide in aqueous
solution.

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00:14:30,000 --> 00:14:34,000
I have a pair of equilibria
that I have written.

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00:14:34,000 --> 00:14:38,000
I have one forward reaction.
And I want to point out,

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00:14:38,000 --> 00:14:43,000
too, that this third equation I
have written here has a special

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name, which is autoprotolysis.

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00:14:52,000 --> 00:14:55,000
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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00:15:19,000 --> 00:15:23,000
pure water.
But at certain points along a

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00:15:23,000 --> 00:15:26,000
titration curve,
such as the one we are going to

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00:15:26,000 --> 00:15:31,000
want to develop on this panel
here, we are going to need to

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00:15:31,000 --> 00:15:36,000
include autoprotolysis in order
to get the right answer because

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00:15:36,000 --> 00:15:40,000
it is there, it can occur.
Let's continue.

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00:16:02,000 --> 00:16:08,000
Another type of equation that
we can use to put this into some

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00:16:08,000 --> 00:16:14,000
kind of a mathematical footing
is the balance of charge in a

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00:16:14,000 --> 00:16:20,000
solution like the one that we
are talking about in that beaker

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00:16:20,000 --> 00:16:25,000
to which we are adding sodium
hydroxide solution.

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00:16:25,000 --> 00:16:31,000
And the charge balance
consideration tells us that the

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00:16:31,000 --> 00:16:37,000
concentration of hydronium ion
plus the concentration of sodium

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00:16:37,000 --> 00:16:43,000
ion must be equal to the
concentration of hydroxide plus

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00:16:43,000 --> 00:16:50,000
the concentration of acetate
ion, A minus.

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00:16:50,000 --> 00:16:53,000
Because what I have done here
in the charge balance

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00:16:53,000 --> 00:16:57,000
consideration is I have said
okay, what are all the possible

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00:16:57,000 --> 00:17:01,000
charged species in solution?
I can make a list of them.

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00:17:01,000 --> 00:17:05,000
It is those four things.
And I know that the number of

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00:17:05,000 --> 00:17:08,000
positive charges must equal the
number of negative charges

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00:17:08,000 --> 00:17:12,000
because this is overall a
neutral beaker in which we are

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00:17:12,000 --> 00:17:15,000
putting things.
But everything that you put in

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00:17:15,000 --> 00:17:18,000
is charge neutral,
so the positives have to equal

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00:17:18,000 --> 00:17:22,000
the negatives.
That is a very nice limiting

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00:17:22,000 --> 00:17:27,000
equation that helps us
understand how the different

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00:17:27,000 --> 00:17:33,000
competing chemical reactions and
equilibria will all settle down

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00:17:33,000 --> 00:17:39,000
and arrive at the physical
result, which will be our

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00:17:39,000 --> 00:17:43,000
observable.
And then, in addition to that,

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00:17:43,000 --> 00:17:46,000
we have a mass balance
consideration.

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00:17:46,000 --> 00:17:50,000
And in this mass balance
consideration,

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00:17:50,000 --> 00:17:57,000
we will be able to write that
our HA initial value --

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00:17:57,000 --> 00:18:01,000
-- that is square brackets,
once again, denoting

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00:18:01,000 --> 00:18:06,000
concentration -- will be,
at any point in time,

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00:18:06,000 --> 00:18:11,000
equal to the actual value of HA
plus the concentration of A

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00:18:11,000 --> 00:18:15,000
minus.
And that just says that when

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00:18:15,000 --> 00:18:20,000
you have acetic acid and you put
it into water,

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00:18:20,000 --> 00:18:25,000
it ionizes partly to hydronium
ion and acetate ion.

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00:18:25,000 --> 00:18:31,000
And that which does not ionize
is unionized HA.

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00:18:31,000 --> 00:18:36,000
And so, the ionized A minus
concentration plus the

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00:18:36,000 --> 00:18:43,000
unionized HA concentration is
equal to the sum total of the

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00:18:43,000 --> 00:18:47,000
acetic acid that is present in
solution.

230
00:18:47,000 --> 00:18:51,000
And, similarly,
we have another mass balance

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00:18:51,000 --> 00:18:57,000
equation, which is just that the
sodium hydroxide initial is

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00:18:57,000 --> 00:19:03,000
equal to our sodium ion
concentration because all of it

233
00:19:03,000 --> 00:19:09,000
is ionized.
That is a little simpler than

234
00:19:09,000 --> 00:19:13,000
the situation with the acetic
acid.

235
00:19:13,000 --> 00:19:20,000
And then, in addition to those
equations, we have our

236
00:19:20,000 --> 00:19:24,000
equilibrium equations.

237
00:19:37,000 --> 00:19:42,000
And we are going to have two of
these because there are two

238
00:19:42,000 --> 00:19:47,000
equilibria that I wrote over
there under titration equations.

239
00:19:47,000 --> 00:19:53,000
The first of these is going to
have the equilibrium constant

240
00:19:53,000 --> 00:19:58,000
Ka, and that is going to be H
three O plus times A

241
00:19:58,000 --> 00:20:02,000
minus,
which is acetate in the case of

242
00:20:02,000 --> 00:20:09,000
acetic acid, all over HA.
That is the expression for the

243
00:20:09,000 --> 00:20:14,000
equilibrium of acetic acid
ionization in dilute aqueous

244
00:20:14,000 --> 00:20:18,000
solution.
And I will give you a value for

245
00:20:18,000 --> 00:20:23,000
this in a moment.
I won't write it right here.

246
00:20:23,000 --> 00:20:28,000
Well, why don't I just do that.
It is 1.8x10^-5 for this

247
00:20:28,000 --> 00:20:35,000
specific example of acetic acid.
And then, if you were working

248
00:20:35,000 --> 00:20:39,000
with some other weak acid,
you would be able to go to a

249
00:20:39,000 --> 00:20:44,000
table and look up the Ka value
for it, if it has been measured

250
00:20:44,000 --> 00:20:48,000
and reported and tabulated.
And then, we have Kw.

251
00:20:48,000 --> 00:20:52,000
And this is the equilibrium
expression for the

252
00:20:52,000 --> 00:20:56,000
autoprotolysis reaction that I
mentioned over there,

253
00:20:56,000 --> 00:21:00,000
the other equilibrium,
which is H three O plus times

254
00:21:00,000 --> 00:21:06,000
OH minus.
And that is equal to 1x10^-14,

255
00:21:06,000 --> 00:21:09,000
so a very small equilibrium
constant.

256
00:21:09,000 --> 00:21:13,000
And some of you may be
wondering why water is not

257
00:21:13,000 --> 00:21:17,000
appearing in an expression like
this one or like this one.

258
00:21:17,000 --> 00:21:22,000
And the reason for this is that
in the expressions that I am

259
00:21:22,000 --> 00:21:26,000
deriving here,
we are using concentrations in

260
00:21:26,000 --> 00:21:30,000
place of activities for these
species.

261
00:21:30,000 --> 00:21:35,000
In the derivation of the
equilibrium constants,

262
00:21:35,000 --> 00:21:40,000
the activity of pure water at
very high concentration is

263
00:21:40,000 --> 00:21:46,000
approximated as one.
And so, this is simplified as

264
00:21:46,000 --> 00:21:50,000
shown here.
And having put all these

265
00:21:50,000 --> 00:21:55,000
equations together,
what we are going to want to do

266
00:21:55,000 --> 00:22:01,000
is take the following approach.
We are looking at the

267
00:22:01,000 --> 00:22:04,000
expression for
electro-neutrality here,

268
00:22:04,000 --> 00:22:08,000
the equal number of positively
and negatively charged species

269
00:22:08,000 --> 00:22:11,000
in solution.
And I want to get this

270
00:22:11,000 --> 00:22:15,000
expression rewritten in terms of
things that I know at any point

271
00:22:15,000 --> 00:22:20,000
along the titration curve,
with the exception of the one

272
00:22:20,000 --> 00:22:22,000
thing that I want to know,
the pH.

273
00:22:22,000 --> 00:22:26,000
And the pH we can calculate
easily if we know the H three O

274
00:22:26,000 --> 00:22:32,000
plus concentration.
So, I am going to be seeking to

275
00:22:32,000 --> 00:22:35,000
get Na plus,
OH minus,

276
00:22:35,000 --> 00:22:39,000
and A minus rewritten
in terms of H three O plus

277
00:22:39,000 --> 00:22:44,000
and constants using
the equations that I have up

278
00:22:44,000 --> 00:22:46,000
here on the board.

279
00:23:00,000 --> 00:23:05,000
First, let's attempt to do that
for A minus using this

280
00:23:05,000 --> 00:23:10,000
Ka expression.
And I am going to rearrange it.

281
00:23:19,000 --> 00:23:27,000
And it is going to be equal to
Ka times HA divided by H three O

282
00:23:27,000 --> 00:23:32,000
plus.
But I still have a problem with

283
00:23:32,000 --> 00:23:37,000
this expression for A minus
because I am going to

284
00:23:37,000 --> 00:23:42,000
need to replace HA with
something that will allow me to

285
00:23:42,000 --> 00:23:47,000
get everything in terms of just
H three O plus and

286
00:23:47,000 --> 00:23:50,000
constants.
And so, if I look over here at

287
00:23:50,000 --> 00:23:55,000
the mass balance equation,
I can use this one and replace

288
00:23:55,000 --> 00:24:01,000
HA with HA initial minus
A minus.

289
00:24:44,000 --> 00:24:47,000
Now I have A minus
written in terms of an initial

290
00:24:47,000 --> 00:24:49,000
concentration of HA itself and
constants.

291
00:24:49,000 --> 00:24:52,000
I need to rearrange this a
little bit.

292
00:24:52,000 --> 00:25:00,000
I am going to multiply through
by H three O plus so

293
00:25:00,000 --> 00:25:08,000
that I can write A minus times H
three O plus is equal to Ka

294
00:25:08,000 --> 00:25:16,000
times HA initial minus Ka times
A minus.

295
00:25:16,000 --> 00:25:24,000
And hopefully I have done that

296
00:25:24,000 --> 00:25:30,000
right.
And, if so, I will be able to

297
00:25:30,000 --> 00:25:39,000
say that A minus is going to be
equal to Ka times HA initial all

298
00:25:39,000 --> 00:25:45,000
over H three O plus.

299
00:25:58,000 --> 00:26:03,000
Does that look good?
Plus Ka, thank you.

300
00:26:03,000 --> 00:26:08,000
H3O+ plus Ka.

301
00:26:08,000 --> 00:26:15,000
And let's bring
that down here.

302
00:26:25,000 --> 00:26:28,000
Because now,
I would like to go ahead and

303
00:26:28,000 --> 00:26:32,000
make this substitution for the
charge.

304
00:26:32,000 --> 00:26:38,000
We have our A minus.
The other ones are a little

305
00:26:38,000 --> 00:26:41,000
easier.
We now have this equation H

306
00:26:41,000 --> 00:26:48,000
three O plus.
And we were going to replace Na

307
00:26:48,000 --> 00:26:53,000
plus by NaOH initial.

308
00:26:53,000 --> 00:27:00,000
And we are going to recognize
that we can use the equation for

309
00:27:00,000 --> 00:27:06,000
autoprotolysis to replace
hydroxide with Kw divided by H

310
00:27:06,000 --> 00:27:14,000
three O plus.
And then now we have the A

311
00:27:14,000 --> 00:27:21,000
minus being replaced
by Ka HA initial all over H

312
00:27:21,000 --> 00:27:29,000
three O plus plus Ka.

313
00:27:29,000 --> 00:27:34,000
And at this point,
you have actually done a really

314
00:27:34,000 --> 00:27:39,000
nice thing, because you have
generated a general titration

315
00:27:39,000 --> 00:27:45,000
equation for the titration of
interest in this particular

316
00:27:45,000 --> 00:27:49,000
case.
You can see that at any point

317
00:27:49,000 --> 00:27:54,000
along that curve for a given
volume m, we can now stop and

318
00:27:54,000 --> 00:28:00,000
walk along in one milliliter
increments.

319
00:28:00,000 --> 00:28:03,000
And, in each spot along that
curve, we could go ahead and

320
00:28:03,000 --> 00:28:07,000
solve this equation to get the
value of H three O plus.

321
00:28:07,000 --> 00:28:09,000
And then we would just have to

322
00:28:09,000 --> 00:28:13,000
take the negative log of that,
and we would have the pH at

323
00:28:13,000 --> 00:28:16,000
that position.
And we are interested if this

324
00:28:16,000 --> 00:28:19,000
type of a model for the
mathematical form of the

325
00:28:19,000 --> 00:28:23,000
titration curve bears any
resemblance to what one sees in

326
00:28:23,000 --> 00:28:27,000
reality.
And, if so, one could then be,

327
00:28:27,000 --> 00:28:31,000
at least for the time being,
until your theory no longer

328
00:28:31,000 --> 00:28:36,000
fits, satisfied that you have
accounted for the various

329
00:28:36,000 --> 00:28:41,000
equilibria that could be present
in a system such as the one we

330
00:28:41,000 --> 00:28:44,000
are describing here.
This happens to be a cubic

331
00:28:44,000 --> 00:28:48,000
equation, and it has,
therefore, three roots.

332
00:28:48,000 --> 00:28:53,000
And when we solve it at each of
these points along the titration

333
00:28:53,000 --> 00:28:57,000
equation, because of the
physically realistic quantities

334
00:28:57,000 --> 00:29:00,000
that we are going to be
interested in,

335
00:29:00,000 --> 00:29:05,000
only one of the roots is
positive.

336
00:29:05,000 --> 00:29:09,000
We are talking about H three O
plus concentration.

337
00:29:09,000 --> 00:29:11,000
It cannot be a negative
concentration.

338
00:29:11,000 --> 00:29:15,000
The positive root is the one we
want, and then we can convert

339
00:29:15,000 --> 00:29:19,000
that to pH and can see what kind
of result we get.

340
00:29:19,000 --> 00:29:22,000
And so, I will show you how we
can do that using a tool

341
00:29:22,000 --> 00:29:25,000
available on Athena.

342
00:30:03,000 --> 00:30:07,000
First of all,
let me tell you that for the

343
00:30:07,000 --> 00:30:13,000
purposes of doing this,
I am going to replace these

344
00:30:13,000 --> 00:30:17,000
various quantities by some
simpler symbols,

345
00:30:17,000 --> 00:30:21,000
here.
I am going to go x plus a,

346
00:30:21,000 --> 00:30:27,000
a is our NaOH initial,
is equal to (b over x) plus (c

347
00:30:27,000 --> 00:30:35,000
times d) over (x plus d).

348
00:30:35,000 --> 00:30:38,000
So, I am going to use those
simpler symbols for this

349
00:30:38,000 --> 00:30:40,000
equation.
I could rearrange it and show

350
00:30:40,000 --> 00:30:43,000
you how it is a cubic equation,
you set it equal to zero,

351
00:30:43,000 --> 00:30:46,000
and then you solve,
and you get the roots.

352
00:30:46,000 --> 00:30:49,000
Just like a quadratic equation
but with another term.

353
00:30:49,000 --> 00:30:52,000
And, instead of doing that,
we are just going to leave it

354
00:30:52,000 --> 00:30:54,000
in this form,
which is simpler to look at.

355
00:30:54,000 --> 00:30:56,000
And we are just going to use
that.

356
00:30:56,000 --> 00:31:00,000
And so, let's go ahead and
define A.

357
00:31:00,000 --> 00:31:03,000
That is the initial
concentration of sodium

358
00:31:03,000 --> 00:31:06,000
hydroxide.
We are going to go 0.1.

359
00:31:06,000 --> 00:31:12,000
And remember I need to get this
in terms of the quantity m here,

360
00:31:12,000 --> 00:31:16,000
which is the volume.
And so, I am going to take m,

361
00:31:16,000 --> 00:31:20,000
minus 20 over m.
And the reason I am doing is

362
00:31:20,000 --> 00:31:26,000
this is I am saying that we have
an 0.1 molar solution of sodium

363
00:31:26,000 --> 00:31:32,000
hydroxide that we are adding.
But there is a dilution factor

364
00:31:32,000 --> 00:31:36,000
that you have to take care of.
So, at every value m we are

365
00:31:36,000 --> 00:31:41,000
saying that we are starting with
a volume of 20 in the acetic

366
00:31:41,000 --> 00:31:44,000
acid flask at the bottom there
at the beginning.

367
00:31:44,000 --> 00:31:48,000
And then we are starting to add
volume units of sodium

368
00:31:48,000 --> 00:31:50,000
hydroxide.
And so the volume,

369
00:31:50,000 --> 00:31:53,000
m, is increasing.
That is how I take care of

370
00:31:53,000 --> 00:31:57,000
that, there.
And then our value of b is Kw,

371
00:31:57,000 --> 00:32:01,000
so that is 1 E minus 14.

372
00:32:01,000 --> 00:32:05,000
That is for autoprotolysis.
And then we have c,

373
00:32:05,000 --> 00:32:09,000
that is our initial
concentration of acetic acid.

374
00:32:09,000 --> 00:32:14,000
And we happen to have a
dilution factor here,

375
00:32:14,000 --> 00:32:18,000
too, but it is simpler,
just 20 divided by m.

376
00:32:18,000 --> 00:32:23,000
And there is that.
And then d, what did I say that

377
00:32:23,000 --> 00:32:26,000
d was?
d is our equilibrium constant

378
00:32:26,000 --> 00:32:31,000
for HA, 1.8 E minus 5.

379
00:32:31,000 --> 00:32:35,000
There is our acidity constant
for acetic acids.

380
00:32:35,000 --> 00:32:40,000
Now I have the four things in
there that I need,

381
00:32:40,000 --> 00:32:46,000
and those are four things that
I know because of the initial

382
00:32:46,000 --> 00:32:51,000
conditions that I am defining
for this problem.

383
00:32:51,000 --> 00:32:57,000
And then I am going to define
my equation as being equal to x

384
00:32:57,000 --> 00:33:03,000
plus a is equal to (b over x)
plus (c times d) all over (x

385
00:33:03,000 --> 00:33:07,000
plus d).
There is my equation.

386
00:33:07,000 --> 00:33:11,000
And this nice calculator,
all I have to do,

387
00:33:11,000 --> 00:33:14,000
in principle,
is tell it what the value of m

388
00:33:14,000 --> 00:33:17,000
is.
And then it is going to solve

389
00:33:17,000 --> 00:33:21,000
that cubic equation,
give me back the three roots.

390
00:33:21,000 --> 00:33:25,000
I take the positive one and I
take the negative log of it,

391
00:33:25,000 --> 00:33:30,000
and then I know the pH at that
value of m.

392
00:33:30,000 --> 00:33:34,000
I can develop a whole titration
curve based on the equations

393
00:33:34,000 --> 00:33:38,000
that we have been discussing.
And so let's choose a value of

394
00:33:38,000 --> 00:33:41,000
m here to start with,
and I am going to choose 30.

395
00:33:41,000 --> 00:33:45,000
We are starting at 20.
At the very beginning of the

396
00:33:45,000 --> 00:33:48,000
titration, we are at 20.
Let's see what happens at 30

397
00:33:48,000 --> 00:33:52,000
given the particular value of m.
Here is what I do.

398
00:33:52,000 --> 00:33:54,000
There may be other ways to do
this.

399
00:33:54,000 --> 00:33:59,000
I am defining this thing called
the solution set.

400
00:33:59,000 --> 00:34:03,000
Because the solving of that
equation is going to give us

401
00:34:03,000 --> 00:34:07,000
three values.
And then I want to be able to

402
00:34:07,000 --> 00:34:12,000
pick the one that I want.
We would use sol-set is equal

403
00:34:12,000 --> 00:34:16,000
to, we are going to solve that
equation, and we want x,

404
00:34:16,000 --> 00:34:21,000
which is our H three O plus
concentration.

405
00:34:21,000 --> 00:34:25,000
And sometimes it chokes.
Let's see what happens.

406
00:34:25,000 --> 00:34:31,000
Sometimes it is a little slow.
But it takes a lot longer if

407
00:34:31,000 --> 00:34:35,000
you actually try to work it all
out by hand.

408
00:34:40,000 --> 00:34:44,000
I have a picture of an image of
this that I grabbed.

409
00:34:44,000 --> 00:34:46,000
I can show you if this fails on
me.

410
00:34:46,000 --> 00:34:49,000
Don't fail on me.
There we go.

411
00:34:49,000 --> 00:34:54,000
You can see it got the answer.
And so here is a positive root,

412
00:34:54,000 --> 00:34:59,000
and then there are two negative
roots to this cubic equation in

413
00:34:59,000 --> 00:35:03,000
x.
And I want to find out the pH.

414
00:35:03,000 --> 00:35:07,000
When we are up to 30 volume
units, m equals 30.

415
00:35:07,000 --> 00:35:13,000
And so, I will define p here as
equal to the negative log base

416
00:35:13,000 --> 00:35:18,000
10 of the first of the answers
in my solution set,

417
00:35:18,000 --> 00:35:23,000
sol-set one,
which is the positive x value.

418
00:35:23,000 --> 00:35:27,000
And, bingo, there is the pH,
4.745 whatever dot,

419
00:35:27,000 --> 00:35:32,000
dot, dot.
That is now the pH at a volume

420
00:35:32,000 --> 00:35:36,000
of 30.
Now, you figure I am starting

421
00:35:36,000 --> 00:35:40,000
with 20 volume units of 0.1
molar acetic acid.

422
00:35:40,000 --> 00:35:43,000
And by the time m equals 30
volume units,

423
00:35:43,000 --> 00:35:49,000
that means I have added ten
volume units of strong base,

424
00:35:49,000 --> 00:35:52,000
enough to quench half of my
acetic acid.

425
00:35:52,000 --> 00:35:57,000
And what if I had just only
quenched none of it?

426
00:35:57,000 --> 00:36:02,000
I can change m now.
Once you have this in here,

427
00:36:02,000 --> 00:36:04,000
you don't have to retype these
things.

428
00:36:04,000 --> 00:36:07,000
I can go back here and say let
m be 20.

429
00:36:07,000 --> 00:36:11,000
And, if I do that,
now I am actually at the zero

430
00:36:11,000 --> 00:36:13,000
point, before I add any strong
base.

431
00:36:13,831 --> 00:00:02,870
And there is the pH now,

432
00:36:16,000 --> 00:36:20,000
The answer to the expression
that we had at the very end of

433
00:36:20,000 --> 00:36:24,000
class last time that was going
to tell us what is the pH of

434
00:36:24,000 --> 00:36:28,000
tenth molar acetic acid,
there is the answer,

435
00:36:28,000 --> 00:36:33,000
2.87, because that is what we
have when we start.

436
00:36:33,000 --> 00:36:40,000
Now you can see that you could
just go through and use that to

437
00:36:40,000 --> 00:36:46,000
get the pH at any value of m
along this titration curve.

438
00:36:46,000 --> 00:36:52,000
And when you do that,
let's show you what the result

439
00:36:52,000 --> 00:36:57,000
is, now what you see here is a
set of data.

440
00:36:57,000 --> 00:37:02,000
I start at 20.
And I'm going up in single

441
00:37:02,000 --> 00:37:04,000
volume units,
21, 22, 23, and 24,

442
00:37:04,000 --> 00:37:09,000
I am showing what the pH is at
each of those choices of m.

443
00:37:09,000 --> 00:37:13,000
And I have gone all the way
from 20 to 60.

444
00:37:13,000 --> 00:37:16,000
And you take that data set --

445
00:37:42,000 --> 00:37:51,000
And then you can take a file,
like the one I just showed you,

446
00:37:51,000 --> 00:38:00,000
which just had values of m and
corresponding values of pH.

447
00:38:00,000 --> 00:38:02,000
And you can go ahead and just
plot that thing.

448
00:38:02,000 --> 00:38:05,000
And, when you do that,
you get the form of the

449
00:38:05,000 --> 00:38:09,000
titration curve here,
which is the classic form of a

450
00:38:09,000 --> 00:38:12,000
titration curve for titrating a
weak acid with a strong base.

451
00:38:12,000 --> 00:38:15,000
And you see that here,
m was 20, and we let it

452
00:38:15,000 --> 00:38:18,000
increment through all the way up
to 60.

453
00:38:18,000 --> 00:38:21,000
And what you see initially,
and you are going to need to

454
00:38:21,000 --> 00:38:25,000
study this one and the other
cases of titration curves that

455
00:38:25,000 --> 00:38:29,000
are in that chapter of your
book, is how the little subtle

456
00:38:29,000 --> 00:38:33,000
features differ from one case to
another.

457
00:38:33,000 --> 00:38:36,000
They do differ.
A much simpler titration curve

458
00:38:36,000 --> 00:38:40,000
than this one is obtained if you
titrate a strong acid with a

459
00:38:40,000 --> 00:38:43,000
strong base.
This is titration of a weak

460
00:38:43,000 --> 00:38:46,000
acid with a strong base.
And you see that initially,

461
00:38:46,000 --> 00:38:50,000
we have that 2.87 initial pH
value, that it kind of shoots up

462
00:38:50,000 --> 00:38:54,000
quickly here at the beginning.
There is a steep rise,

463
00:38:54,000 --> 00:38:57,000
initially.
And then it kind of levels off.

464
00:38:57,000 --> 00:39:01,000
And in this region here,
the pH is not changing very

465
00:39:01,000 --> 00:39:06,000
rapidly at all as you are adding
more sodium hydroxide.

466
00:39:06,000 --> 00:39:09,000
Until you get out here,
and the pH just shoots up.

467
00:39:09,000 --> 00:39:13,000
And you will notice that it is
around where we get to 40.

468
00:39:13,000 --> 00:39:18,000
Because we started with 20 mLs
of tenth molar acetic acid.

469
00:39:18,000 --> 00:39:22,000
And, when we get out to m
equals 40, we have added 20 mLs

470
00:39:22,000 --> 00:39:26,000
of tenth molar sodium hydroxide.
At that point the pH just

471
00:39:26,000 --> 00:39:29,000
shoots up.
And then over here it turns

472
00:39:29,000 --> 00:39:33,000
over and just goes kind of
slowly for a minute.

473
00:39:33,000 --> 00:39:37,000
And it just approaches an
asymptotic limiting value over

474
00:39:37,000 --> 00:39:39,000
here on the high pH side of the
scale.

475
00:39:39,000 --> 00:39:44,000
One thing you will notice here
is that at this point where we

476
00:39:44,000 --> 00:39:48,000
have added the same number of
equivalence of sodium hydroxide

477
00:39:48,000 --> 00:39:52,000
molecules as we had acetic acid
molecules to start with,

478
00:39:52,000 --> 00:39:56,000
this occurs over here at
greater than 8.0 pH units rather

479
00:39:56,000 --> 00:00:07,000
than at the neutral value of

480
00:40:00,000 --> 00:40:03,000
And I would like you to think
about why that might be.

481
00:40:03,000 --> 00:40:08,000
And that will be something that
will be covered in recitation,

482
00:40:08,000 --> 00:40:12,000
but it is important.
If we had derived here today

483
00:40:12,000 --> 00:40:16,000
the expression for the titration
of a strong acid with a strong

484
00:40:16,000 --> 00:40:19,000
base, then at 40 we would be
right at 7.0,

485
00:40:19,000 --> 00:40:23,000
but when we were using this
weak acid, acetic acid,

486
00:40:23,000 --> 00:40:27,000
it comes up here over 8.0 pH
units.

487
00:40:27,000 --> 00:40:31,000
This point where we have added
as many strong base molecules as

488
00:40:31,000 --> 00:40:33,000
we started with,
weak acid molecules,

489
00:40:33,000 --> 00:40:38,000
is called the equivalence or
the stoichiometric point in the

490
00:40:38,000 --> 00:40:40,000
titration.
Moreover, where would your

491
00:40:40,000 --> 00:40:43,000
indicator molecule change color?
At what point,

492
00:40:43,000 --> 00:40:45,000
at what value of m,
for example?

493
00:40:45,000 --> 00:40:49,000
It would be close to there.
It would be because,

494
00:40:49,000 --> 00:40:53,000
in the case of phenolphthalein,
it changes between right in

495
00:40:53,000 --> 00:40:55,000
here.
It might change right around

496
00:40:55,000 --> 00:41:00,000
maybe 41 or so.
m equals 41 might be where the

497
00:41:00,000 --> 00:41:04,000
color change from colorless to
pink occurs, for the reasons

498
00:41:04,000 --> 00:41:07,000
that we have discussed.
And then right here,

499
00:41:07,000 --> 00:41:11,000
it turns out at this value of m
equals 30, the pH is equal to

500
00:41:11,000 --> 00:41:14,000
the pKa of acetic acid.
And that is a relationship that

501
00:41:14,000 --> 00:41:19,000
I would like you to think about.
And it is a relationship that

502
00:41:19,000 --> 00:41:22,000
comes into play very strongly
when we talk about the

503
00:41:22,000 --> 00:41:25,000
preparation of buffer solutions
according to the

504
00:41:25,000 --> 00:41:30,000
Henderson-Hasselbach equation.
At m equals 30,

505
00:41:30,000 --> 00:41:34,000
the pH value of the titration
is the pKa of acetic acid,

506
00:41:34,000 --> 00:41:37,000
which is right here at about
4.7 or so.

507
00:41:37,000 --> 00:41:41,000
We did that one a moment ago.
We found out what that value

508
00:41:41,000 --> 00:41:43,000
was.
And, in plus or minus 1.0 pH

509
00:41:43,000 --> 00:41:46,000
units of that,
we have the part of the

510
00:41:46,000 --> 00:41:50,000
titration curve that is called
the buffering region.

511
00:41:50,000 --> 00:41:53,000
In here we have generated a
buffer in situ.

512
00:41:53,000 --> 00:41:57,000
And buffers help the pH not to
change too much when acids or

513
00:41:57,000 --> 00:42:02,000
bases are added in solution.
That is what buffers do.

514
00:42:02,000 --> 00:42:06,000
They try to help you keep in a
constant pH range because lots

515
00:42:06,000 --> 00:42:10,000
of things are dependent upon a
particular pH for proper

516
00:42:10,000 --> 00:42:12,000
functioning.
And you know that this is

517
00:42:12,000 --> 00:42:16,000
important in the world's oceans,
it is important in our

518
00:42:16,000 --> 00:42:19,000
environment, and it is important
in our bodies.

519
00:42:19,000 --> 00:42:23,000
So, the pH considerations for
aqueous solutions are quite

520
00:42:23,000 --> 00:42:26,000
important.
And I hope that today I have

521
00:42:26,000 --> 00:42:31,000
been able to give you a nice way
of thinking about some of the

522
00:42:31,000 --> 00:42:33,000
math that goes behind a
titration curve.

523
00:42:33,691 --> 00:42:36,000
Have a nice Halloween.