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PROFESSOR: Please settle
down and take a look

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at this question.
 

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OK, let's take 10 seconds.
 

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I think that it's a simple
math mistake is between

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one and two at least.
 

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So, the trick here is you know
the p h and the p k a and you

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want to find the ratio so you
can subtract and do the log.

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So maybe we'll have this
question later or something

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similar and we can
try this one again.

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So we're going to talk
about buffers again today.

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I just feel the need to take a
moment and reflect on the

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historic events of the last 24
hours, and talk about how

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it will affect chemistry.
 

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So some of you may have
voted for the first time.

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Some of you may have worked on
a campaign for the first time.

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Some of you may have been very
active in a campaign for the

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first time, either for Obama or
McCain, that you got involved.

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And I thought just to put this
election in a little bit of the

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historic perspective in terms
of about being an undergraduate

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student or a student and
working on a political

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campaign or being part of
a political movement.

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So, my father was very
active as a political

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student activist.
 

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But the difference between some
of you and my father was that

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he was a political activist at
the University of Hamberg in

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Germany in the 1930's
in Hitler's Germany.

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So he was the leader of the
left wing student organization.

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That was something that put
one's life at risk to take

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on that role at that time.
 

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So, things were heating up a
little bit and the gestapo

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were discussing some of the
activities with the left wing

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student leaders at college
campuses in Germany.

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And some of them, after the
discussions, no one knew

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where they went, they
seemed to disappear.

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Now my father was very
concerned about this and he

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decided to lay low for a while,
and so he thought I'll do a

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semester at another university.
 

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And he told his parents that if
the gestapo came looking for

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him, that they should send him
a telegram saying "Your Aunt

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Millie is sick." Since he did
not have an Aunt Milly, he

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knew that that would
mean get out now.

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So he went to another
university and he was doing a

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semester there, and someone he
knew told him you really

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need to go into hiding.
 

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But he didn't really trust this
person, so we packed a bag with

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a few clothes and some
toiletries, but he

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didn't actually leave.
 

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Then the next day he came home
and there was a telegram

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under his door.
 

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So, you can guess what
the telegram said.

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He grabbed the bag that
was already packed and

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headed down the stairs.
 

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The gestapo was coming
up the stairs.

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My father's name was Heinz
Leopold Lushinski and the

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gestapo said to him, "Do
you know Herr Lushinski?

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And my father said, "Yes, of
course, he lives on the top

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floor." The gestapo went up, my
father went down, and he didn't

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go back to Germany
for 30 years.

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So he came to the United
States as a political refugee

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and became a citizen.
 

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He voted in every election,
every possibility, he

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was very, very active.
 

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My family was very, very
active in politics.

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He gave money every year to the
American Civil Liberties Union

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to protect civil liberties, and
he also gave money to the

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American Rifle Association.
 

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He always liked to
have a plan b.

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So, it was sometimes a little
humbling to be the only

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child of this man.
 

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He was in his 50's when I was
born, and I thought how can I

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live up to something like this?
 

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Am I ever going to risk my
life for what I believe in.

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If given that choice would
I do the right thing?

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And I don't know if I'll ever
get an answer to that question,

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but I talked to my father about
this and he said all I need to

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do is work hard, find something
that I love doing, some way

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that I can contribute, and
that's what's really important

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-- contributing is
really important.

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So I was drawn to teaching, and
I love teaching here at MIT

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because you all are so talented
and smart, and it is really an

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honor and a privilege to be
involved in your education.

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But I feel that in the last 24
hours, we have all received an

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additional call to service.
 

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That president elect Obama said
in the campaign that his top

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priorities are going to be
scientific research, coming up

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with clean energy technologies,
and improving healthcare.

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He called to scientists
and engineers.

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And last night the American
people said yes, we like

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that vision, and they
elected him president.

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So we have been called, you
have been called, he has

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reached out to students and
said, students of science

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and engineering, you
need to contribute.

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And it's been a while since any
president has really called to

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action, scientists
and engineers.

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And last time that happened,
a man went on the moon.

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So let's see what we
can do this time.

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The next challenge is
clean energy, healthcare.

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It's going to be really
important for sciences and

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engineers to get involved,
and at the core an energy

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technologies, and at the core
of medicine is chemistry.

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So you are in the right
place right now.

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You are going to be the
generation that needs to solve

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these problems, because if you
don't solve the energy problem

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and don't come up with clean
alternatives, there isn't going

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to be much of a planet left for
another generation to try

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to solve those problems.
 

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So it's going to be your job
and your job is starting

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right now with the education
that you can get at MIT.

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So, it's actually somewhat
interesting that today, the day

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after this election, we are
going to talk about one of the

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units that students in this
class have had the most

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difficulty with over the
years, acid based titrations.

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This has been the undoing of
some chemistry individuals.

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It has been the undoing
of some grades of A.

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It has been the undoing,
perhaps, of some

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interest in chemistry.
 

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But I would like to say today,
at this moment, it will not be

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your undoing, it will
be your triumph.

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Every year I challenge students
to do the best job on acid

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based titration ever, and
people have been doing well.

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This might be the last
time I teach in the fall.

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You have actually had the
highest grades so far in this

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class, in the history of the
class that I know of, and

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so this is the challenge.
 

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So right after this election,
your challenge is to conquer

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chemistry starting one acid
and one base at a time.

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So, ready to do some
acid based titrations?

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Who are the naysayers
in this crowd?

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Just a few people up there.
 

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All right.
 

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I have to tell you that what
I'm going to tell you about

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acid based titrations will seem
like it makes pretty good

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sense as I'm saying it.
 

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But often, people inform me
that when they actually go to

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work the problems on the test,
it's a little less clear on

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what they're supposed
to be doing.

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So the key to acid based
titrations is really

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to work problems.
 

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And so we have, for your
benefit, assigned problems for

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the problem-set due Friday.
 

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And so after today, you should
be set to do all of the

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problems on the problem-set.
 

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And in terms of acid based
titration, you will need a lot

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of this knowledge again in
organic chemistry,

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biochemistry, if you go to
medical school -- I used to TA

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medical students, they didn't
know how to do this.

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And I said "Who taught you
freshmen chemistry?" So it's

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good to learn to this now here
today, work problems, take the

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next test, and guaranteed
it'll be on the final again.

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So you'll learn it now, you'll
get lots of points, both on

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the final and the third exam.
 

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All right, so acid based
titrations, they're not that

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hard, but there are not a lot
of equations to use, and

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I think that people in
chemistry are used to

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what equation do I use.
 

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No, it's really about thinking
about what's going on in the

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problem, and as the problem
proceeds, as more, say, strong

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base is added, the
problem changes.

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So it's figuring out where
you are in the titration

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and knowing what sort
of steps to apply.

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So here are some titration
curves, and one thing you may

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be asked to do is draw a
titration curve, so you

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should be familiar with
what they look like.

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So we talked last time about
strong acids and strong bases.

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So if you have a strong base,
you're going to have a basic p

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h, and then as you add the
strong acid, you will go to the

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equivalence point, equivalence
point when you've added the

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same amount of moles of acid as
there is base or base as there

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is acid, equal number of moles.
 

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And when you mix a strong acid
in a strong base, you form a

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salt, and the salt is neutral
in p h, because the conjugate

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of a strong acid or a strong
base, is ineffectual, it

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doesn't affect the
p h, it's neutral.

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So we have p h 7, and then you
continue to add, in this case,

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a more strong acid, and
the p h goes down.

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So for the other titration it's
pretty much the same, except

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you start at acidic p h's, go
up to neutral p h,

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and then go basic.
 

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So we talked about these last
time and we worked a couple of

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problems, but now we're going
to move into the slightly more

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difficult type of problem,
which has to do with when you

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have a weak acid or a weak
base being titrated.

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So let's look at the difference
of the curve to start off with.

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So here we have the strong acid
and the strong base, and here

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we have a weak acid
and a strong base.

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One thing you may notice right
off is that the equivalence

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point has a different p h.
 

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So, a strong acid and strong
base again, mix, you form a

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salt that's neutral, p h 7.
 

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But if you're titrating a weak
acid in a strong base, the

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conjugate of the strong base
will be ineffective, but

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the conjugate of the weak
acid will act as a base.

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So the p h then, at the
equivalence point, when you've

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added equal number of moles of
your strong base as you had

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weak acid, then you'll have the
conjugate base around, and the

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p h will be greater than 7.
 

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So in working the problems, if
you get an answer with this

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type of titration problem
that's different than that for

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p h at the equivalence point,
you're going to know that you

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did something wrong, you need
to go back and check your math.

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Another big difference has to
do with the curve shape down

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here, and so you notice a
difference over here

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than over there.
 

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And in a titration that
involves a weak acid in a

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strong base, you have a part of
the curve that's known as a

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buffering region, and the p h
is fairly flat in this

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buffering region as
shown down here.

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So that's in contrast,
there's no such buffering

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region on this side.
 

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So here the p h will go
up, it'll level off,

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and then go up again.
 

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And this, for some of you, is
probably the frustration in

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doing acid based titrations in
lab, because you're adding and

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nothing's happening and
nothing's happening and

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nothing's happening, and you're
in this region, then all of a

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sudden you add just a little
more and you're up here.

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So notice how steep
that is over here.

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So sometimes when you're in the
buffering region, it seems like

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you're never going to reach the
end of the titration and then

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it'll happen all too quickly.
 

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So buffering region, remember a
buffer is something that has a

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conjugate, weak acid and weak
base pair, and then in a

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buffering region, the p h
pretty much stays fairly

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constant in that region.
 

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It acts as a buffer,
neutralizing the p h,

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maintaining the p h by being a
source or sink of protons, and

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so here the p h then is
staying constant in

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that buffering region.
 

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So those are some of the
differences between

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the type of curves.
 

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Another point that I will
mention or term I will mention

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that has to do with weak acid
in strong base or a weak base

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in strong acid is this 1/2
equivalence point concept.

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So 1/2 equivalence point you've
added 1/2 of the amount of

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strong base that you need to
get to the equivalence point,

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and that's right in the middle
of that buffering region.

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So that's another point
where you'll be asked

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to calculate the p h.
 

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So now let's look at different
points in a titration.

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So, first let's walk through
and just think about

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what is happening.
 

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So when we start in this
titration of a weak acid in a

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00:14:43 --> 00:14:48
strong base, before we've added
any of the strong base, all

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we have is a weak acid.
 

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So it is a weak acid in
water type problem.

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And so here I've drawn our
acid, and the acid has its

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00:14:57 --> 00:15:00
proton, which is going to
give up when you start

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00:15:00 --> 00:15:03
doing the titration.
 

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So that's what we
have at zero volume.

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Then we start adding our strong
base, and the strong base is

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00:15:10 --> 00:15:13
going to react with the acid,
one-to-one stoichiometry,

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it's a strong base.
 

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It'll pull off protons off the
same number of moles of the

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strong acid as the number
of moles of the strong

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00:15:23 --> 00:15:24
base that were added.
 

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And so then, you'll start
to have a mixture of your

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conjugates, your weak acid
and your conjugate base.

266
00:15:31 --> 00:15:33
So the base is a minus here.
 

267
00:15:33 --> 00:15:36
And so if you have a mixture of
a weak acid in its conjugate

268
00:15:36 --> 00:15:39
base, that's a buffer, and
so you'll move in to the

269
00:15:39 --> 00:15:40
buffering region here.
 

270
00:15:40 --> 00:15:45
So that's at any volume that is
greater than zero and less than

271
00:15:45 --> 00:15:52
the equivalence point is going
to be around in that region.

272
00:15:52 --> 00:15:55
Then we have a special category
of the buffering region, which

273
00:15:55 --> 00:15:57
is when you've added the
volume to get to the

274
00:15:57 --> 00:15:59
1/2 equivalence point.
 

275
00:15:59 --> 00:16:03
And when you've done that, you
will have converted 1/2 of the

276
00:16:03 --> 00:16:06
weak acid it its conjugate
base, so you'll have equal

277
00:16:06 --> 00:16:11
number of moles of your weak
acid as moles of the conjugate

278
00:16:11 --> 00:16:14
base -- 1/2 has been converted.
 

279
00:16:14 --> 00:16:18
And so that's a special
category right there.

280
00:16:18 --> 00:16:20
Then you get to the
equivalence point.

281
00:16:20 --> 00:16:22
At the equivalence point,
you've added the same number of

282
00:16:22 --> 00:16:26
moles of strong base as the
number of moles of weak acid

283
00:16:26 --> 00:16:30
you have, so you've converted
all of your weak acid to

284
00:16:30 --> 00:16:31
it's conjugate base.
 

285
00:16:31 --> 00:16:34
So all you have is conjugate
base now, and so that's

286
00:16:34 --> 00:16:38
controlling the p h, so the p
h should be greater than 7.

287
00:16:38 --> 00:16:42
So that's a weak base
in water problem.

288
00:16:42 --> 00:16:48
And if you keep going, then
you're going to end up with a

289
00:16:48 --> 00:16:50
strong base in water problem.
 

290
00:16:50 --> 00:16:55
The weak base will still be
around, but it will be

291
00:16:55 --> 00:16:58
negligibly affecting the p h
compared to the fact that

292
00:16:58 --> 00:17:00
you're dumping strong acid
into your titration.

293
00:17:00 --> 00:17:03
And so that's this
part of the curve.

294
00:17:03 --> 00:17:09
So you see that in one type of
problem, one titration problem,

295
00:17:09 --> 00:17:13
you actually have a lot of sub
problems, or sub types of

296
00:17:13 --> 00:17:17
problems, you'll have weak acid
buffer, special category of

297
00:17:17 --> 00:17:21
buffer, a conjugate base or a
salt issue, and then

298
00:17:21 --> 00:17:22
a strong base.
 

299
00:17:22 --> 00:17:24
And this is one of the things
that people have trouble with

300
00:17:24 --> 00:17:27
in the titrations, because we
may not ask you to do all the

301
00:17:27 --> 00:17:30
points, we may just sort of
jump in somewhere, and say

302
00:17:30 --> 00:17:32
okay, what is the p h at the
equivalenced point, and you

303
00:17:32 --> 00:17:35
need to think about what's
happened to get to the

304
00:17:35 --> 00:17:36
equivalence point.
 

305
00:17:36 --> 00:17:39
Or we may jump in and ask you
about a region that would be in

306
00:17:39 --> 00:17:42
the buffering region, and you
have to remember that at that

307
00:17:42 --> 00:17:45
point you should have some of
the weak acid and also some of

308
00:17:45 --> 00:17:47
the conjugate bases
being formed.

309
00:17:47 --> 00:17:50
So, it seems like there are a
lot of different things, but

310
00:17:50 --> 00:17:52
there are only five
types of problems.

311
00:17:52 --> 00:17:56
But in a titration curve, you
run into a lot of those

312
00:17:56 --> 00:18:02
different types at different
points in the problem.

313
00:18:02 --> 00:18:04
So now let's go the other
direction and consider

314
00:18:04 --> 00:18:08
titration of a weak base
with a strong acid.

315
00:18:08 --> 00:18:10
So here's what that
curve would look like.

316
00:18:10 --> 00:18:13
You're going to start basic, of
course, because you're starting

317
00:18:13 --> 00:18:16
with a weak base, you haven't
added any strong acid yet.

318
00:18:16 --> 00:18:20
As you add strong acid,
the p h will decrease.

319
00:18:20 --> 00:18:23
Because it is a weak base, you
will be forming some of its

320
00:18:23 --> 00:18:27
conjugate as you add the strong
acid, and so you'll go through

321
00:18:27 --> 00:18:30
a buffering region again where
the curve would be flat, where

322
00:18:30 --> 00:18:33
the p h will be pretty much
the same for region of time.

323
00:18:33 --> 00:18:36
Then the curve will drop
again and you'll get to

324
00:18:36 --> 00:18:38
the equivalence point.
 

325
00:18:38 --> 00:18:42
At the equivalence point,
you've added the same amount of

326
00:18:42 --> 00:18:46
moles of strong acid as you had
weak base, so all of your weak

327
00:18:46 --> 00:18:49
base is converted to its
conjugate acid, and so you

328
00:18:49 --> 00:18:53
should be acidic at the
equivalence point, and

329
00:18:53 --> 00:18:55
then the curve goes down.
 

330
00:18:55 --> 00:18:58
So again, we can think
about this in terms

331
00:18:58 --> 00:18:59
of what is happening.
 

332
00:18:59 --> 00:19:03
In the beginning it's just a
weak base in water problem, but

333
00:19:03 --> 00:19:07
as you add strong acid, you
were pronating some of your

334
00:19:07 --> 00:19:11
base and forming its conjugate
acid here, and you're in the

335
00:19:11 --> 00:19:13
going to be in the
buffering region.

336
00:19:13 --> 00:19:17
Then at the 1/2 equivalence
point, you've added enough

337
00:19:17 --> 00:19:21
moles of strong acid to convert
1/2 of the weak base to its

338
00:19:21 --> 00:19:24
conjugate, so those are going
to be equal to each other --

339
00:19:24 --> 00:19:27
the number of moles of the weak
base and the number of moles

340
00:19:27 --> 00:19:29
of its conjugate acid.
 

341
00:19:29 --> 00:19:32
At the equivalence point,
you've converted all of the

342
00:19:32 --> 00:19:35
weak base you started with to
its conjugate acid, so it'll be

343
00:19:35 --> 00:19:39
a weak acid in water problem,
and then at the end

344
00:19:39 --> 00:19:41
it's strong acid.
 

345
00:19:41 --> 00:19:45
So the trick is to recognizing
what type of problem you're

346
00:19:45 --> 00:19:49
being asked to do, and a lot of
times if people get a question

347
00:19:49 --> 00:19:52
and they just write down OK, at
this point in the titration

348
00:19:52 --> 00:19:55
curve, it's going to be a
weak base in water problem.

349
00:19:55 --> 00:19:58
And just writing that down,
most of the time if you get

350
00:19:58 --> 00:20:01
that far, you do the rest
of the problem correctly.

351
00:20:01 --> 00:20:05
So just identifying the type,
there are only 5, of problems

352
00:20:05 --> 00:20:12
gets you a long way to
getting the right answer.

353
00:20:12 --> 00:20:16
So let's do an example.
 

354
00:20:16 --> 00:20:19
We're going to titrate a weak
acid with a strong base.

355
00:20:19 --> 00:20:25
We have 25 mils of 0.1 molar
acid with 0.15 moles of a

356
00:20:25 --> 00:20:33
strong base, n a o h, we're
given the k a for the acid.

357
00:20:33 --> 00:20:39
First we start with 0 mils
of the strong base added.

358
00:20:39 --> 00:20:45
So what type of
problem is this?

359
00:20:45 --> 00:20:48
It's a weak acid problem.
 

360
00:20:48 --> 00:20:51
So we know how to write the
equation for a weak acid

361
00:20:51 --> 00:20:52
or for an acid in water.
 

362
00:20:52 --> 00:20:56
We have the acid in water
going to hydronium ions

363
00:20:56 --> 00:21:01
and a conjugate base.
 

364
00:21:01 --> 00:21:03
So weak acid.
 

365
00:21:03 --> 00:21:10
For weak acid, we're going
to use our k a, and we're

366
00:21:10 --> 00:21:13
going to set up our
equilibrium expression.

367
00:21:13 --> 00:21:18
So here we have 0.1
molar of our acid.

368
00:21:18 --> 00:21:21
We're going to have some of
that go away in the

369
00:21:21 --> 00:21:26
equilibrium, forming hydronium
ion and some conjugate base,

370
00:21:26 --> 00:21:30
and so we know we have
expressions for the

371
00:21:30 --> 00:21:32
concentrations at equilibrium.
 

372
00:21:32 --> 00:21:36
And we can use our k a, k a
for acid, it's a weak acid

373
00:21:36 --> 00:21:42
problem, and we can look at
products over reactants.

374
00:21:42 --> 00:21:44
So, see, now we're doing a
titration problem, but you

375
00:21:44 --> 00:21:47
already know how to do this
problem because we've seen a

376
00:21:47 --> 00:21:50
weak acid in water
problem before.

377
00:21:50 --> 00:21:56
So we have x squared
over 0.10 minus x here.

378
00:21:56 --> 00:22:03
We can assume x is small, and
get rid of this minus x, and

379
00:22:03 --> 00:22:05
then later go back and check
it, so that just makes the

380
00:22:05 --> 00:22:07
math a little bit easier.
 

381
00:22:07 --> 00:22:14
And we can solve for x and then
we can check -- we can take

382
00:22:14 --> 00:22:21
this value, 0.00421 over 0.1
and see whether that's less

383
00:22:21 --> 00:22:24
than 5%, it's close but it is.
 

384
00:22:24 --> 00:22:26
So that assumption is OK.
 

385
00:22:26 --> 00:22:30
If it wasn't, what
would we have to do?

386
00:22:30 --> 00:22:32
Quadratic equation.
 

387
00:22:32 --> 00:22:35
All right, so now, here's
a sig fig question.

388
00:22:35 --> 00:23:20
Tell me how many sig figs
this p h actually has.

389
00:23:20 --> 00:23:42
OK, 10 seconds.
 

390
00:23:42 --> 00:23:46
So, in the first part of the
problem we had a concentration

391
00:23:46 --> 00:23:51
that had 2 significant
figures, the 0.10 molar.

392
00:23:51 --> 00:23:54
Sometimes later, people have
extra significant figures that

393
00:23:54 --> 00:23:57
they're carrying along, but we
had those 2, and so we're

394
00:23:57 --> 00:24:01
going to have 2 after the
decimal point then in

395
00:24:01 --> 00:24:05
the answer of the p h.
 

396
00:24:05 --> 00:24:07
So again, the number of
significant figures that are

397
00:24:07 --> 00:24:15
limiting are going to be the
number after the decimal point.

398
00:24:15 --> 00:24:20
All right, so we have
one p h value, and now

399
00:24:20 --> 00:24:21
we're going to move on.
 

400
00:24:21 --> 00:24:26
So let me just put our
one p h value down.

401
00:24:26 --> 00:24:37
We have volume of strong base,
and p h over here, and we're

402
00:24:37 --> 00:24:40
starting here with
zero moles added.

403
00:24:40 --> 00:24:43
We have a p h of 2 .
 

404
00:24:43 --> 00:24:43
38.
 

405
00:24:43 --> 00:24:49
It's a weak acid, so it should
be an acidic p h, which it is.

406
00:24:49 --> 00:24:54
All right, so now let's
move into the titration

407
00:24:54 --> 00:24:56
problem, and now 5 .
 

408
00:24:56 --> 00:25:02
0 mils of the strong base have
been added, and we need to

409
00:25:02 --> 00:25:04
find what the p h is now.
 

410
00:25:04 --> 00:25:08
So it's a strong base,
so it's going to react

411
00:25:08 --> 00:25:10
almost completely,
that's our assumption.

412
00:25:10 --> 00:25:13
If it's strong, it
goes completely.

413
00:25:13 --> 00:25:18
And so, the number of moles of
the strong base that we add

414
00:25:18 --> 00:25:22
will convert all of the same
number of moles of our acid

415
00:25:22 --> 00:25:25
over to its conjugate.
 

416
00:25:25 --> 00:25:29
So we can just do a
subtraction then.

417
00:25:29 --> 00:25:31
So first, we need to know
the initial moles of

418
00:25:31 --> 00:25:33
the acid that we had.
 

419
00:25:33 --> 00:25:36
We had 25 mils, 0.10 molar.
 

420
00:25:36 --> 00:25:40
We calculate the number of
moles for the hydroxide added,

421
00:25:40 --> 00:25:44
we added 5 mils, it was 0.15
molar, and so we can calculate

422
00:25:44 --> 00:25:48
the number of moles of the
strong base that were added.

423
00:25:48 --> 00:25:53
So the strong base will react
completely with the same number

424
00:25:53 --> 00:25:55
of moles of the weak acid.
 

425
00:25:55 --> 00:25:58
And we're going to do then --
we have the moles of the weak

426
00:25:58 --> 00:26:01
acid here, minus the number of
moles of the strong base we've

427
00:26:01 --> 00:26:03
added, and so we're
going to have 1 .

428
00:26:03 --> 00:26:09
75 times 10 to the minus 3
moles of the weak acid left.

429
00:26:09 --> 00:26:14
So, then how many moles of
the conjugate base will be

430
00:26:14 --> 00:26:19
formed by this reaction?
 

431
00:26:19 --> 00:26:28
What do you think?
 

432
00:26:28 --> 00:26:29
Same number.
 

433
00:26:29 --> 00:26:39
So 0.75 times 10
to the minus 3.

434
00:26:39 --> 00:26:42
So always remember that in
these titration problems,

435
00:26:42 --> 00:26:46
nothing has been added yet,
you're at zero mils added.

436
00:26:46 --> 00:26:48
Some amount of some
subtractions are going to

437
00:26:48 --> 00:26:50
have to occur because
something has happened.

438
00:26:50 --> 00:26:53
You've converted something,
things are different

439
00:26:53 --> 00:26:54
than when you started.
 

440
00:26:54 --> 00:27:01
All right, so now we have weak
acid and we have moles of its

441
00:27:01 --> 00:27:07
conjugate, what type
of problem is this?

442
00:27:07 --> 00:27:09
If you have a weak acid
and its conjugate base

443
00:27:09 --> 00:27:16
-- buffer, right.
 

444
00:27:16 --> 00:27:19
So we're going to do a buffer
problem and we need to

445
00:27:19 --> 00:27:21
know the molarity first.
 

446
00:27:21 --> 00:27:26
So we have moles over volume --
again, the volume, you had 25

447
00:27:26 --> 00:27:29
mils to begin with,
you added 5 more.

448
00:27:29 --> 00:27:32
So you have to have the total
volume 30 mils, and we

449
00:27:32 --> 00:27:37
can calculate then the
concentrations of both.

450
00:27:37 --> 00:27:40
Now we can set up our
equilibrium table, and this

451
00:27:40 --> 00:27:44
looks like a buffer problem
because it is, and by looking

452
00:27:44 --> 00:27:46
like a buffer problem you
something over here, you have

453
00:27:46 --> 00:27:49
your weak acid over here, but
you have something over here

454
00:27:49 --> 00:27:52
now, it's not zero now, we're
starting with some

455
00:27:52 --> 00:27:53
conjugate base.
 

456
00:27:53 --> 00:28:01
So we have 0.0583 minus x on
one side, and we 0.025 molar

457
00:28:01 --> 00:28:05
plus x on the other side.
 

458
00:28:05 --> 00:28:07
We can use k a again.
 

459
00:28:07 --> 00:28:12
This is set up as an acid in
water going to hydronium ions

460
00:28:12 --> 00:28:18
and conjugate base, so we can
use our k a, set things up,

461
00:28:18 --> 00:28:22
and we can always say let's
see if x is small, make an

462
00:28:22 --> 00:28:24
assumption, check it later.
 

463
00:28:24 --> 00:28:27
That'll simplify the math.
 

464
00:28:27 --> 00:28:30
So we get rid of the
plus x and the minus x.

465
00:28:30 --> 00:28:33
Again, we're saying that if x
is small, the initial

466
00:28:33 --> 00:28:35
concentrations are going to be
more or less the same as the

467
00:28:35 --> 00:28:40
concentrations after the
equilibration occurs.

468
00:28:40 --> 00:28:42
And we can calculate 4 .
 

469
00:28:42 --> 00:28:46
13 times 10 to the minus
4, as x, that is a

470
00:28:46 --> 00:28:48
pretty small number.
 

471
00:28:48 --> 00:28:51
And we have to check it, and
yup, it's small enough, it's

472
00:28:51 --> 00:28:56
under 5%, so that's OK.
 

473
00:28:56 --> 00:28:58
So now we can plug this in.
 

474
00:28:58 --> 00:29:01
X is our hydronium ion
concentration minus log of the

475
00:29:01 --> 00:29:05
hydronium ion concentration
is p h, and we can

476
00:29:05 --> 00:29:07
calculate p h to 3 .
 

477
00:29:07 --> 00:29:12
38 -- again, we're limited
by two significant figures

478
00:29:12 --> 00:29:14
in the concentration.
 

479
00:29:14 --> 00:29:19
So now we've added 5 mils down
here, and our p h has gone up

480
00:29:19 --> 00:29:24
a little bit, it's now at 3 .
 

481
00:29:24 --> 00:29:30
38 over here.
 

482
00:29:30 --> 00:29:33
There's another option
for a buffer problem.

483
00:29:33 --> 00:29:39
What's the one equation
in this unit?

484
00:29:39 --> 00:29:43
Our friend Henderson
Hasselbalch.

485
00:29:43 --> 00:29:47
And yes, you can use that here
too, assuming that you check

486
00:29:47 --> 00:29:49
the assumption and it's OK.
 

487
00:29:49 --> 00:29:52
Most people will prefer
to do this because

488
00:29:52 --> 00:29:54
it is a bit easier.
 

489
00:29:54 --> 00:29:58
So, you weren't given, though,
the p k a in this problem, you

490
00:29:58 --> 00:30:02
were given the k a, so pretty
easy to calculate -- minus

491
00:30:02 --> 00:30:05
log of the k a is the p k a.
 

492
00:30:05 --> 00:30:08
So you can calculate
that, put that in.

493
00:30:08 --> 00:30:10
You have your concentrations
and it should be

494
00:30:10 --> 00:30:13
concentrations, but you may
notice that if you actually

495
00:30:13 --> 00:30:16
had moles the volume
would cancel here.

496
00:30:16 --> 00:30:20
So here are the concentrations,
but with the same volume,

497
00:30:20 --> 00:30:22
the volume term does cancel.
 

498
00:30:22 --> 00:30:25
It makes this a little faster
and it gives the same

499
00:30:25 --> 00:30:27
answer, which is great.
 

500
00:30:27 --> 00:30:30
To use Henderson Hasselbalch
you also need the 5% rule to

501
00:30:30 --> 00:30:33
be true, because Henderson
Hasselbalch is assuming

502
00:30:33 --> 00:30:35
that x is small.
 

503
00:30:35 --> 00:30:37
It's assuming that the initial
concentrations and the

504
00:30:37 --> 00:30:42
concentrations after
equilibrium are about the same.

505
00:30:42 --> 00:30:44
So we can check the assumption.
 

506
00:30:44 --> 00:30:47
We can back-calculate the
hydronium ion concentration,

507
00:30:47 --> 00:30:49
which would be x, and see
if it's small, we already

508
00:30:49 --> 00:30:52
know it is, so it's OK.
 

509
00:30:52 --> 00:30:56
So there are 2 options for
buffer problems, but do not use

510
00:30:56 --> 00:30:58
the Henderson Hasselbalch
equation when it isn't in the

511
00:30:58 --> 00:31:02
buffering region, it
doesn't hold then.

512
00:31:02 --> 00:31:04
So again, you check the
assumption, and if

513
00:31:04 --> 00:31:06
it's OK, it's fine.
 

514
00:31:06 --> 00:31:09
If not, you need to use option
one and you need to use

515
00:31:09 --> 00:31:14
the quadratic equation.
 

516
00:31:14 --> 00:31:18
All right, so buffering region.
 

517
00:31:18 --> 00:31:21
Now we're at the special
kind of problem in the

518
00:31:21 --> 00:31:25
buffering region, the
1/2 equivalence point.

519
00:31:25 --> 00:31:30
So here you've added 1/2 the
number of moles of the strong

520
00:31:30 --> 00:31:33
base to convert 1/2 the
moles of the weak acid

521
00:31:33 --> 00:31:35
to its conjugate.
 

522
00:31:35 --> 00:31:40
So at this point, the
concentration of h a equals the

523
00:31:40 --> 00:31:44
concentration of a minus --
equal number of moles in the

524
00:31:44 --> 00:31:47
same volume, those are equal.
 

525
00:31:47 --> 00:31:51
You can use Henderson
Hasselbalch here, and find that

526
00:31:51 --> 00:31:55
if they're equal, you're
talking about minus log of 1,

527
00:31:55 --> 00:32:02
so the p h is going
to equal the p k a.

528
00:32:02 --> 00:32:05
And you're done with
this type of problem.

529
00:32:05 --> 00:32:09
I have been known to put 1/2
equivalence problems on an

530
00:32:09 --> 00:32:13
exam, because exams are often
long, you have only 50 minutes,

531
00:32:13 --> 00:32:16
there's lots of different type
of problems, and this problem

532
00:32:16 --> 00:32:19
should not take you a
long amount of time.

533
00:32:19 --> 00:32:22
You do not have to prove
to me that this is true.

534
00:32:22 --> 00:32:26
All you need to remember, 1/2
equivalence point, p h equals

535
00:32:26 --> 00:32:32
p k a, and if you calculate
the p k a, you're done.

536
00:32:32 --> 00:32:35
So this is a short
type of problem.

537
00:32:35 --> 00:32:37
If you remember the definition
of 1/2 equivalence

538
00:32:37 --> 00:32:40
point, it's easy to do.
 

539
00:32:40 --> 00:32:45
So now we have another
number, so 3 .

540
00:32:45 --> 00:32:58
75, and we're working
on our curve.

541
00:32:58 --> 00:33:01
Now let's move to the
equivalence point.

542
00:33:01 --> 00:33:05
At the equivalence point,
you've added the same number

543
00:33:05 --> 00:33:09
of moles of your strong
base as you had weak acid.

544
00:33:09 --> 00:33:13
So you've converted all
of your weak acid to

545
00:33:13 --> 00:33:18
its conjugate base.
 

546
00:33:18 --> 00:33:20
So the p h should
be greater than 7.

547
00:33:20 --> 00:33:24
Now all you have is conjugate
base, that's basic, p h

548
00:33:24 --> 00:33:28
should be greater than 7.
 

549
00:33:28 --> 00:33:33
So when you are doing this
titration, you have your weak

550
00:33:33 --> 00:33:35
acid and your strong base.
 

551
00:33:35 --> 00:33:40
You're going to be forming a
salt here, and a salt problem,

552
00:33:40 --> 00:33:43
you can tell me about salts.
 

553
00:33:43 --> 00:33:49
And so, just remind me, what
does the n a plus contribute

554
00:33:49 --> 00:33:54
to the p h here.
 

555
00:33:54 --> 00:33:55
It's going to be neutral.
 

556
00:33:55 --> 00:34:00
And what about this
guy down here?

557
00:34:00 --> 00:34:02
Yeah, so it's going
to be basic.

558
00:34:02 --> 00:34:06
So, the sodium, anything group
1, group 2, no effect on

559
00:34:06 --> 00:34:07
p h, they're neutral.
 

560
00:34:07 --> 00:34:10
But if you have a conjugate
base of a weak acid,

561
00:34:10 --> 00:34:13
that's going to be basic.
 

562
00:34:13 --> 00:34:22
Salt problems, really just part
of what you already know about.

563
00:34:22 --> 00:34:24
So always check your work.
 

564
00:34:24 --> 00:34:27
If your p h doesn't make sense
from what you know, you might

565
00:34:27 --> 00:34:31
have made a math mistake.
 

566
00:34:31 --> 00:34:35
So let's calculate the actual
p h at the equivalence point.

567
00:34:35 --> 00:34:37
We know that it should
be basic, but what

568
00:34:37 --> 00:34:41
is it going to be?
 

569
00:34:41 --> 00:34:45
So first, we need to know how
much of the strong base we had

570
00:34:45 --> 00:34:50
to add, because we need to
know about all the moles.

571
00:34:50 --> 00:34:53
So how much of this
did we need to add.

572
00:34:53 --> 00:34:56
So we needed to add enough of
the strong base that you

573
00:34:56 --> 00:34:59
converted all of the moles of
the weak acid to its conjugate.

574
00:34:59 --> 00:35:00
So we had 2 .
 

575
00:35:00 --> 00:35:04
5 times 10 to the minus 3
moles of our weak acid.

576
00:35:04 --> 00:35:07
So that's all going to be
converted to the moles of the

577
00:35:07 --> 00:35:10
conjugate base, and so that's
going to be equal to the number

578
00:35:10 --> 00:35:12
of moles we needed to do it.
 

579
00:35:12 --> 00:35:13
So we needed 2 .
 

580
00:35:13 --> 00:35:17
5 times 10 to the minus 3 moles
of our strong base to do

581
00:35:17 --> 00:35:19
that complete conversion.
 

582
00:35:19 --> 00:35:22
We know the concentration
of the base was 0.15 .

583
00:35:22 --> 00:35:23
So we would have needed 1 .
 

584
00:35:23 --> 00:35:28
67 times 10 to the minus 2
liters of this concentration

585
00:35:28 --> 00:35:32
added to reach the
equivalence point.

586
00:35:32 --> 00:35:34
So then the total volume that
we're going to have at the

587
00:35:34 --> 00:35:38
equivalence point is the 25
mils that we had to begin

588
00:35:38 --> 00:35:40
with, plus this 16 .
 

589
00:35:40 --> 00:35:45
7 mils to make this
final, total volume.

590
00:35:45 --> 00:35:48
And remember, you always need
to think, what is the total

591
00:35:48 --> 00:35:51
volume, how much has been
added to get to this point

592
00:35:51 --> 00:35:54
in the titration curve.
 

593
00:35:54 --> 00:35:58
Then we can calculate molarity,
so we know how many moles of

594
00:35:58 --> 00:36:00
conjugate base have been
formed, and we know the new

595
00:36:00 --> 00:36:05
volume, so we can calculate the
concentration of the

596
00:36:05 --> 00:36:08
conjugate base.
 

597
00:36:08 --> 00:36:12
So now, you can help me
solve this problem.

598
00:36:12 --> 00:36:43
Set up an equation
for me to solve it.

599
00:36:43 --> 00:37:02
Let's take 10 seconds.
 

600
00:37:02 --> 00:37:05
That's the best score
we've had today.

601
00:37:05 --> 00:37:06
Yup.
 

602
00:37:06 --> 00:37:11
So now we're talking
about a conjugate base.

603
00:37:11 --> 00:37:16
So we have converted all of the
weak acid to the conjugate

604
00:37:16 --> 00:37:23
base, and so it's a weak base
in water problem, so we're

605
00:37:23 --> 00:37:26
going to talk about a k b.
 

606
00:37:26 --> 00:37:29
If you were only given the
k a for this problem, how

607
00:37:29 --> 00:37:35
would you find k b -- what
interconnects k a and k b?

608
00:37:35 --> 00:37:37
K w, right.
 

609
00:37:37 --> 00:37:40
So you can calculate, here it's
given to you, but you could

610
00:37:40 --> 00:37:43
calculate it if you had a
calculator, and you would

611
00:37:43 --> 00:37:44
find that this is true.
 

612
00:37:44 --> 00:37:46
Now it's a weak base
in water problem.

613
00:37:46 --> 00:37:48
We're not in the buffering
region anymore.

614
00:37:48 --> 00:37:51
We've converted all of our
weak acid to the conjugate.

615
00:37:51 --> 00:37:54
So it's a weak base
in water problem.

616
00:37:54 --> 00:37:59
So we have x squared, 0.06,
that was the concentration

617
00:37:59 --> 00:38:04
we calculated, minus x.
 

618
00:38:04 --> 00:38:08
So again, think about what
type of problem it is.

619
00:38:08 --> 00:38:12
So again, weak base in
water problem -- x squared

620
00:38:12 --> 00:38:16
over 0.06 minus x.
 

621
00:38:16 --> 00:38:22
And we can assume that x is
small, and calculate a value

622
00:38:22 --> 00:38:27
for x, which is 0.83 times
10 to the minus 6, and then

623
00:38:27 --> 00:38:31
we're going to calculate p
o h, because now x is the

624
00:38:31 --> 00:38:33
hydroxide ion concentration.
 

625
00:38:33 --> 00:38:39
Because in a weak base in water
problem, here in this type of

626
00:38:39 --> 00:38:42
problem, the base, and here is
your acid -- the conjugate of

627
00:38:42 --> 00:38:46
this acid is the base,
hydroxide, and the conjugate of

628
00:38:46 --> 00:38:51
this weak base is its conjugate
acid over here, so now when we

629
00:38:51 --> 00:38:54
are solving for x, we're
solving for hydroxide ion in

630
00:38:54 --> 00:38:58
concentration, so we're
calculating a p o h, which then

631
00:38:58 --> 00:39:02
we can calculate a p h from.
 

632
00:39:02 --> 00:39:05
So we can take 14 minus 5 .
 

633
00:39:05 --> 00:39:08
74 and get our value.
 

634
00:39:08 --> 00:39:12
And it's bigger than neutral,
it's 8, it's basic, and that

635
00:39:12 --> 00:39:17
makes sense, it is a weak
base in water problem.

636
00:39:17 --> 00:39:19
So, let's see, it's 8 .
 

637
00:39:19 --> 00:39:26
26, so now we're up here in
our curve, and we're at 8 .

638
00:39:26 --> 00:39:31
26, and that's going to
be greater than 7 for

639
00:39:31 --> 00:39:34
this type of problem.
 

640
00:39:34 --> 00:39:38
So that makes sense, it's good.
 

641
00:39:38 --> 00:39:44
Greater than 7 is
what we want to see.

642
00:39:44 --> 00:39:48
So now, you've gone too far --
you've passed the equivalence

643
00:39:48 --> 00:39:55
point, and you keep adding
your strong base in.

644
00:39:55 --> 00:39:59
Now you still have some of the
weak conjugate base around.

645
00:39:59 --> 00:40:02
So you still have this around,
but you only have 1 .

646
00:40:02 --> 00:40:05
83 times 10 to the
minus 6 molar of it.

647
00:40:05 --> 00:40:08
So very little amount
-- x is small.

648
00:40:08 --> 00:40:13
So your p h is going to be
dictated by the amount of extra

649
00:40:13 --> 00:40:18
strong base you're adding.
 

650
00:40:18 --> 00:40:21
So this is similar, then,
to a strong acid or strong

651
00:40:21 --> 00:40:25
base in water problem.
 

652
00:40:25 --> 00:40:30
So if you're 5 mils past the
equivalence point, 5 mils times

653
00:40:30 --> 00:40:34
your concentration of a strong
base, so you have extra, 7 .

654
00:40:34 --> 00:40:38
5 times 10 to the
minus 4 moles extra.

655
00:40:38 --> 00:40:42
So then you need to calculate a
concentration of that, and so

656
00:40:42 --> 00:40:46
you remember the whole volume
-- you're 5 mils past, you had

657
00:40:46 --> 00:40:49
25 miles to start with,
and you had to add 16 .

658
00:40:49 --> 00:40:52
7 mils to get to the
equivalence point.

659
00:40:52 --> 00:40:54
And you have, that's your
total volume, you get a

660
00:40:54 --> 00:40:58
concentration, that's your
concentration of hydroxide,

661
00:40:58 --> 00:41:00
it reacts completely,
you don't have to do any

662
00:41:00 --> 00:41:02
equilibrium table here.
 

663
00:41:02 --> 00:41:04
It's going complete,
it's a strong base.

664
00:41:04 --> 00:41:09
You could try adding that value
of your other weak base to

665
00:41:09 --> 00:41:13
this, but remember, that's
times 10 to the minus 6, so

666
00:41:13 --> 00:41:15
it's not going to be
significant with

667
00:41:15 --> 00:41:16
significant figures.
 

668
00:41:16 --> 00:41:19
So you can just use this
value -- plug it in to p

669
00:41:19 --> 00:41:23
o h, calculate it, and
then calculate p h.

670
00:41:23 --> 00:41:32
And so now we're somewhere
up here at p h 12 .

671
00:41:32 --> 00:41:39
21, 5 mils past.
 

672
00:41:39 --> 00:41:46
And there we've worked
a titration problem.

673
00:41:46 --> 00:41:49
So let's review what we saw.
 

674
00:41:49 --> 00:41:52
In the beginning, zero mils of
the strong base, we have a

675
00:41:52 --> 00:41:54
weak acid in water problem.
 

676
00:41:54 --> 00:41:57
We moved into the buffering
region where we had our weak

677
00:41:57 --> 00:42:01
acid and the conjugate
base of that weak acid.

678
00:42:01 --> 00:42:04
At the equivalence point, we've
converted all of the weak acid

679
00:42:04 --> 00:42:07
to the conjugate base, so
it's a weak base problem.

680
00:42:07 --> 00:42:10
And then beyond the equivalence
point, it's a strong

681
00:42:10 --> 00:42:12
base problem.
 

682
00:42:12 --> 00:42:18
That's what we've just worked.
 

683
00:42:18 --> 00:42:20
So, we can check
these all off now.

684
00:42:20 --> 00:42:24
You know how to do all of
these types of problems.

685
00:42:24 --> 00:42:27
And there are not that many,
you just need to figure

686
00:42:27 --> 00:42:30
out where to apply what.
 

687
00:42:30 --> 00:42:33
And if you can do that, you're
all set, this unit will be

688
00:42:33 --> 00:42:37
easy for you, and you can
go through and make me

689
00:42:37 --> 00:42:39
very happy on the exam.
 

690
00:42:39 --> 00:42:41
There's nothing -- well, there
are few things in life as

691
00:42:41 --> 00:42:45
beautiful to me as a perfectly
worked titration problem.

692
00:42:45 --> 00:42:49
It really, it brings me joy,
and I've had people write on

693
00:42:49 --> 00:42:54
the exam sometimes, "I hope
that my solution to this brings

694
00:42:54 --> 00:42:57
you joy." And I will often
write, "Yes, it does,"

695
00:42:57 --> 00:42:59
and put a smiley face.
 

696
00:42:59 --> 00:43:02
Because it really is nice to
see these beautifully worked.

697
00:43:02 --> 00:43:08
I know, I'm a little nerdy and
geeky, but after yesterday,

698
00:43:08 --> 00:43:16
being smart and a nerd and
a geek is cool again.

699
00:43:16 --> 00:43:20
All right, so let me just
tell you where we're going.

700
00:43:20 --> 00:43:23
We have five more minutes, and
actually that's perfect,

701
00:43:23 --> 00:43:29
because I can get through some
rules in those 5 minutes.

702
00:43:29 --> 00:43:31
So let's do 5 minutes of rules.
 

703
00:43:31 --> 00:43:35
Oxidation reduction doesn't
have a lot of rules, so five

704
00:43:35 --> 00:43:38
minutes is actually all
we need to do that.

705
00:43:38 --> 00:43:42
Oxidation reduction involves
equilibrium, it involves

706
00:43:42 --> 00:43:43
thermodynamics.
 

707
00:43:43 --> 00:43:46
I like it because it's really
important for reactions

708
00:43:46 --> 00:43:50
occurring in the body, and acid
bases as well -- p k a's are

709
00:43:50 --> 00:43:52
really important to that.
 

710
00:43:52 --> 00:43:55
And so, between acid base and
oxidation reduction, you cover

711
00:43:55 --> 00:43:59
the way a lot of enzymes work.
 

712
00:43:59 --> 00:44:01
So let me give you five minutes
of rules, and that will

713
00:44:01 --> 00:44:03
serve you well in this unit.
 

714
00:44:03 --> 00:44:06
Some of these are
pretty simple.

715
00:44:06 --> 00:44:10
For free elements, each atom
has an oxidation number

716
00:44:10 --> 00:44:15
of 0, so this would be 0.
 

717
00:44:15 --> 00:44:19
So, oxidation number of
0 in a free element.

718
00:44:19 --> 00:44:25
For ions that are composed of
one atom, the oxidation number

719
00:44:25 --> 00:44:30
is equal to the charge of the
atom, so lithium plus 1 ions

720
00:44:30 --> 00:44:34
would have an oxidation
number of plus 1.

721
00:44:34 --> 00:44:37
Again, pretty straightforward.
 

722
00:44:37 --> 00:44:40
Group one and group two
make your lives easy.

723
00:44:40 --> 00:44:42
They seem to have a lot
of consistent rules.

724
00:44:42 --> 00:44:44
Group one metals in the
periodic table have

725
00:44:44 --> 00:44:47
oxidation numbers of 1.
 

726
00:44:47 --> 00:44:50
Group two metals have
oxidation numbers of plus 2.

727
00:44:50 --> 00:44:55
Aluminum is plus 3 in
all its compounds.

728
00:44:55 --> 00:44:57
Pretty simple.
 

729
00:44:57 --> 00:45:00
Now we get to things that are
a little more complicated

730
00:45:00 --> 00:45:02
but still useful, oxygen.
 

731
00:45:02 --> 00:45:09
Oxygen is mostly minus 2, but
there are exceptions to that,

732
00:45:09 --> 00:45:14
such as in peroxides where it
can have an oxidation number of

733
00:45:14 --> 00:45:21
minus 1, and if it's with a
group one metal, it

734
00:45:21 --> 00:45:23
can be minus 1.
 

735
00:45:23 --> 00:45:28
Remember, group one, and
actually group two here, that's

736
00:45:28 --> 00:45:32
plus 1, always plus 1, always
plus 1, always plus 2, and

737
00:45:32 --> 00:45:35
so hydrogen has to
accommodate that.

738
00:45:35 --> 00:45:39
So usually plus 1, except when
it's in a binary complex with

739
00:45:39 --> 00:45:44
these particular metals that
are in group one or group two.

740
00:45:44 --> 00:45:49
Fluorine, almost always minus
1, or always minus 1 -- other

741
00:45:49 --> 00:45:55
halogens, a chloride, bromide,
iodide, also usually negatives,

742
00:45:55 --> 00:46:00
but if they're with
oxygen, then it changes.

743
00:46:00 --> 00:46:04
So, here is an example.
 

744
00:46:04 --> 00:46:07
And in neutral molecules,
the sum of the oxidation

745
00:46:07 --> 00:46:09
numbers must be 0.
 

746
00:46:09 --> 00:46:12
When the molecule has a charge,
the sum of the oxidation

747
00:46:12 --> 00:46:18
numbers must be equal
to that charge.

748
00:46:18 --> 00:46:21
So, let's do a quick example.
 

749
00:46:21 --> 00:46:28
Hydrogen, in this case,
is going to be what?

750
00:46:28 --> 00:46:34
Plus 1, so it's not with
a group one, group

751
00:46:34 --> 00:46:35
two metal here.
 

752
00:46:35 --> 00:46:40
So what does that
leave for nitrogen?

753
00:46:40 --> 00:46:46
And that makes the sum, plus 1,
which is equal to the sum of

754
00:46:46 --> 00:46:49
that molecule, so that works.
 

755
00:46:49 --> 00:46:52
So we might not have known
nitrogen, but we can figure it

756
00:46:52 --> 00:46:55
out if we know the rules for
hydrogen and we know what

757
00:46:55 --> 00:46:57
it all has to equal up to.
 

758
00:46:57 --> 00:46:59
And so, this unit is sometimes
a relief after oxidation

759
00:46:59 --> 00:47:02
reduction, because it's all
about simple adding and

760
00:47:02 --> 00:47:05
subtracting, it's not so bad.
 

761
00:47:05 --> 00:47:08
OK, oxidation numbers do
not have to be integers.

762
00:47:08 --> 00:47:12
Example here, you have
superoxide, what would

763
00:47:12 --> 00:47:16
its oxidation number be?
 

764
00:47:16 --> 00:47:19
Minus 1/2.
 

765
00:47:19 --> 00:47:24
And those are the rules, and
then on Friday, we'll come

766
00:47:24 --> 00:47:27
back and we'll look
at some examples.

767
00:47:27 --> 00:47:28