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PROFESSOR: Please settle
down and take a

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00:00:23,460 --> 00:00:45,790
look at this question.

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00:00:45,790 --> 00:01:10,170
OK, let's take 10 seconds.

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00:01:10,170 --> 00:01:16,630
I think that it's a simple math
mistake is between one

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and two at least. So, the trick
here is you know the p h

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00:01:21,170 --> 00:01:24,880
and the p k a and you want to
find the ratio so you can

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00:01:24,880 --> 00:01:27,350
subtract and do the log.

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00:01:27,350 --> 00:01:31,000
So maybe we'll have this
question later or something

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00:01:31,000 --> 00:01:33,910
similar and we can try
this one again.

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00:01:33,910 --> 00:01:37,490
So we're going to talk about
buffers again today.

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00:01:37,490 --> 00:01:43,360
I just feel the need to take
a moment and reflect on the

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00:01:43,360 --> 00:01:48,200
historic events of the last 24
hours, and talk about how it

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00:01:48,200 --> 00:01:55,130
will affect chemistry.

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00:01:55,130 --> 00:01:59,400
So some of you may have voted
for the first time.

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00:01:59,400 --> 00:02:02,640
Some of you may have worked on a
campaign for the first time.

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00:02:02,640 --> 00:02:06,650
Some of you may have been very
active in a campaign for the

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00:02:06,650 --> 00:02:10,050
first time, either for
Obama or McCain,

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00:02:10,050 --> 00:02:12,640
that you got involved.

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00:02:12,640 --> 00:02:15,250
And I thought just to put this
election in a little bit of

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00:02:15,250 --> 00:02:18,270
the historic perspective in
terms of about being an

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00:02:18,270 --> 00:02:21,560
undergraduate student or a
student and working on a

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00:02:21,560 --> 00:02:25,610
political campaign or being part
of a political movement.

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00:02:25,610 --> 00:02:30,110
So, my father was very active
as a political student

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00:02:30,110 --> 00:02:33,830
activist. But the difference
between some of you and my

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00:02:33,830 --> 00:02:37,090
father was that he was a
political activist at the

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00:02:37,090 --> 00:02:40,970
University of Hamberg in Germany
in the 1930's in

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00:02:40,970 --> 00:02:42,780
Hitler's Germany.

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00:02:42,780 --> 00:02:45,370
So he was the leader of
the left wing student

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organization.

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00:02:47,430 --> 00:02:52,630
That was something that put
one's life at risk to take on

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00:02:52,630 --> 00:02:55,690
that role at that time.

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00:02:55,690 --> 00:02:59,050
So, things were heating up a
little bit and the gestapo

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00:02:59,050 --> 00:03:02,370
were discussing some of the
activities with the left wing

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00:03:02,370 --> 00:03:06,210
student leaders at college
campuses in Germany.

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00:03:06,210 --> 00:03:09,930
And some of them, after the
discussions, no one knew where

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they went, they seemed
to disappear.

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00:03:11,980 --> 00:03:14,590
Now my father was very concerned
about this and he

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00:03:14,590 --> 00:03:17,130
decided to lay low for a while,
and so he thought I'll

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00:03:17,130 --> 00:03:19,320
do a semester at another
university.

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00:03:19,320 --> 00:03:21,970
And he told his parents that
if the gestapo came looking

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

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Aunt 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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00:03:32,740 --> 00:03:34,920
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 need

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to go into hiding.

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

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bag with a few clothes and some
toiletries, but he didn't

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actually leave. Then the next
day he came home and there was

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a telegram 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 headed down

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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

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didn't 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 was

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very, very active.

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00:04:28,460 --> 00:04:31,120
My family was very, very
active in politics.

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He gave money every year to the
American Civil Liberties

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Union to protect civil
liberties, and he also gave

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money to the 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 child

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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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00:05:00,820 --> 00:05:03,720
And I don't know if I'll ever
get an answer to that

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00:05:03,720 --> 00:05:07,210
question, but I talked to my
father about this and he said

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all I need to do is work hard,
find something that I love

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00:05:10,940 --> 00:05:14,020
doing, some way that I can
contribute, and that's what's

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00:05:14,020 --> 00:05:15,040
really important --

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00:05:15,040 --> 00:05:17,300
contributing is really important.

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00:05:17,300 --> 00:05:21,770
So I was drawn to teaching, and
I love teaching here at

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MIT because you all are so
talented and smart, and it is

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00:05:26,230 --> 00:05:29,160
really an honor and a privilege
to be involved in

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00:05:29,160 --> 00:05:30,970
your education.

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00:05:30,970 --> 00:05:36,010
But I feel that in the last 24
hours, we have all received an

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00:05:36,010 --> 00:05:39,010
additional call to service.

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00:05:39,010 --> 00:05:44,910
That president elect Obama said
in the campaign that his

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00:05:44,910 --> 00:05:48,310
top priorities are going to be
scientific research, coming up

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00:05:48,310 --> 00:05:51,420
with clean energy technologies,
and improving

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healthcare.

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00:05:52,390 --> 00:05:55,780
He called to scientists
and engineers.

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00:05:55,780 --> 00:05:59,120
And last night the American
people said yes, we like that

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00:05:59,120 --> 00:06:01,870
vision, and they elected
him president.

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00:06:01,870 --> 00:06:05,000
So we have been called, you
have been called, he has

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00:06:05,000 --> 00:06:08,500
reached out to students and
said, students of science and

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00:06:08,500 --> 00:06:10,880
engineering, you need
to contribute.

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00:06:10,880 --> 00:06:13,590
And it's been a while since
any president has really

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00:06:13,590 --> 00:06:16,160
called to action, scientists
and engineers.

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00:06:16,160 --> 00:06:19,380
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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00:06:21,890 --> 00:06:25,210
The next challenge is clean
energy, healthcare.

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00:06:25,210 --> 00:06:27,220
It's going to be really
important for sciences and

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00:06:27,220 --> 00:06:31,460
engineers to get involved,
and at the core an energy

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00:06:31,460 --> 00:06:38,270
technologies, and at the core
of medicine is chemistry.

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00:06:38,270 --> 00:06:41,450
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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00:06:44,430 --> 00:06:48,360
these problems, because if you
don't solve the energy problem

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00:06:48,360 --> 00:06:51,850
and don't come up with clean
alternatives, there isn't

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00:06:51,850 --> 00:06:54,800
going to be much of a planet
left for another generation to

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00:06:54,800 --> 00:06:58,910
try to solve those problems. So
it's going to be your job

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00:06:58,910 --> 00:07:02,270
and your job is starting right
now with the education that

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00:07:02,270 --> 00:07:05,830
you can get at MIT.

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00:07:05,830 --> 00:07:09,770
So, it's actually somewhat
interesting that today, the

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00:07:09,770 --> 00:07:13,300
day after this election, we are
going to talk about one of

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00:07:13,300 --> 00:07:17,070
the units that students in this
class have had the most

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00:07:17,070 --> 00:07:22,930
difficulty with over the years,
acid based titrations.

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00:07:22,930 --> 00:07:27,760
This has been the undoing of
some chemistry individuals.

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00:07:27,760 --> 00:07:31,440
It has been the undoing of some
grades of A. It has been

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00:07:31,440 --> 00:07:35,580
the undoing, perhaps, of some
interest in chemistry.

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00:07:35,580 --> 00:07:39,600
But I would like to say today,
at this moment, it will not be

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00:07:39,600 --> 00:07:44,910
your undoing, it will
be your triumph.

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00:07:44,910 --> 00:07:48,310
Every year I challenge students
to do the best job on

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00:07:48,310 --> 00:07:52,350
acid based titration ever, and
people have been doing well.

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00:07:52,350 --> 00:07:55,090
This might be the last time
I teach in the fall.

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00:07:55,090 --> 00:07:58,320
You have actually had the
highest grades so far in this

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00:07:58,320 --> 00:08:02,160
class, in the history of the
class that I know of, and so

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00:08:02,160 --> 00:08:03,740
this is the challenge.

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00:08:03,740 --> 00:08:07,600
So right after this election,
your challenge is to conquer

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00:08:07,600 --> 00:08:18,570
chemistry starting one acid
and one base at a time.

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00:08:18,570 --> 00:08:21,730
So, ready to do some acid
based titrations?

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00:08:21,730 --> 00:08:31,360
Who are the naysayers
in this crowd?

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00:08:31,360 --> 00:08:33,050
Just a few people up there.

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00:08:33,050 --> 00:08:46,310
All right.

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00:08:46,310 --> 00:08:49,380
I have to tell you that what
I'm going to tell you about

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00:08:49,380 --> 00:08:52,650
acid based titrations will seem
like it makes pretty good

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00:08:52,650 --> 00:08:54,520
sense as I'm saying it.

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00:08:54,520 --> 00:08:57,740
But often, people inform me that
when they actually go to

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00:08:57,740 --> 00:09:01,750
work the problems on the test,
it's a little less clear on

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00:09:01,750 --> 00:09:03,390
what they're supposed
to be doing.

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00:09:03,390 --> 00:09:06,750
So the key to acid based
titrations is really to work

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00:09:06,750 --> 00:09:11,100
problems. And so we have, for
your benefit, assigned

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00:09:11,100 --> 00:09:13,470
problems for the problem-set
due Friday.

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00:09:13,470 --> 00:09:15,800
And so after today, you should
be set to do all of the

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00:09:15,800 --> 00:09:18,070
problems on the problem-set.

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00:09:18,070 --> 00:09:21,490
And in terms of acid based
titration, you will need a lot

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00:09:21,490 --> 00:09:23,900
of this knowledge again
in organic chemistry,

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00:09:23,900 --> 00:09:26,180
biochemistry, if you go
to medical school --

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00:09:26,180 --> 00:09:27,910
I used to TA medical students,
they didn't

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00:09:27,910 --> 00:09:29,280
know how to do this.

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00:09:29,280 --> 00:09:32,450
And I said "Who taught you
freshmen chemistry?" So it's

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00:09:32,450 --> 00:09:38,240
good to learn to this now here
today, work problems, take the

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00:09:38,240 --> 00:09:42,190
next test, and guaranteed it'll
be on the final again.

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00:09:42,190 --> 00:09:45,490
So you'll learn it now, you'll
get lots of points, both on

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00:09:45,490 --> 00:09:49,810
the final and the third exam.

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00:09:49,810 --> 00:09:54,390
All right, so acid based
titrations, they're not that

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00:09:54,390 --> 00:09:58,300
hard, but there are not a lot
of equations to use, and I

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00:09:58,300 --> 00:10:00,460
think that people in chemistry
are used to what

165
00:10:00,460 --> 00:10:02,140
equation do I use.

166
00:10:02,140 --> 00:10:04,600
No, it's really about thinking
about what's going on in the

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00:10:04,600 --> 00:10:08,270
problem, and as the problem
proceeds, as more, say, strong

168
00:10:08,270 --> 00:10:10,680
base is added, the
problem changes.

169
00:10:10,680 --> 00:10:13,010
So it's figuring out where you
are in the titration and

170
00:10:13,010 --> 00:10:17,800
knowing what sort of
steps to apply.

171
00:10:17,800 --> 00:10:20,640
So here are some titration
curves, and one thing you may

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00:10:20,640 --> 00:10:24,440
be asked to do is draw a
titration curve, so you should

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00:10:24,440 --> 00:10:27,360
be familiar with what
they look like.

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00:10:27,360 --> 00:10:32,330
So we talked last time about
strong acids and strong bases.

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00:10:32,330 --> 00:10:38,210
So if you have a strong base,
you're going to have a basic p

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00:10:38,210 --> 00:10:42,720
h, and then as you add the
strong acid, you will go to

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00:10:42,720 --> 00:10:45,430
the equivalence point,
equivalence point when you've

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00:10:45,430 --> 00:10:49,520
added the same amount of moles
of acid as there is base or

179
00:10:49,520 --> 00:10:52,560
base as there is acid, equal
number of moles.

180
00:10:52,560 --> 00:10:55,970
And when you mix a strong acid
in a strong base, you form a

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00:10:55,970 --> 00:10:59,880
salt, and the salt is neutral in
p h, because the conjugate

182
00:10:59,880 --> 00:11:02,700
of a strong acid or a strong
base, is ineffectual, it

183
00:11:02,700 --> 00:11:04,830
doesn't affect the p
h, it's neutral.

184
00:11:04,830 --> 00:11:09,030
So we have p h 7, and then you
continue to add, in this case,

185
00:11:09,030 --> 00:11:11,580
a more strong acid, and
the p h goes down.

186
00:11:11,580 --> 00:11:13,970
So for the other titration it's
pretty much the same,

187
00:11:13,970 --> 00:11:18,150
except you start at acidic p
h's, go up to neutral p h, and

188
00:11:18,150 --> 00:11:20,100
then go basic.

189
00:11:20,100 --> 00:11:23,460
So we talked about these last
time and we worked a couple of

190
00:11:23,460 --> 00:11:26,360
problems, but now we're going to
move into the slightly more

191
00:11:26,360 --> 00:11:30,180
difficult type of problem, which
has to do with when you

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00:11:30,180 --> 00:11:34,400
have a weak acid or a weak
base being titrated.

193
00:11:34,400 --> 00:11:36,620
So let's look at the difference
of the curve to

194
00:11:36,620 --> 00:11:38,190
start off with.

195
00:11:38,190 --> 00:11:41,500
So here we have the strong acid
and the strong base, and

196
00:11:41,500 --> 00:11:44,120
here we have a weak acid
and a strong base.

197
00:11:44,120 --> 00:11:47,530
One thing you may notice right
off is that the equivalence

198
00:11:47,530 --> 00:11:50,220
point has a different p h.

199
00:11:50,220 --> 00:11:53,440
So, a strong acid and strong
base again, mix, you form a

200
00:11:53,440 --> 00:11:56,410
salt that's neutral, p h 7.

201
00:11:56,410 --> 00:12:00,755
But if you're titrating a weak
acid in a strong base, the

202
00:12:00,755 --> 00:12:03,520
conjugate of the strong base
will be ineffective, but the

203
00:12:03,520 --> 00:12:07,450
conjugate of the weak acid
will act as a base.

204
00:12:07,450 --> 00:12:10,460
So the p h then, at the
equivalence point, when you've

205
00:12:10,460 --> 00:12:14,730
added equal number of moles of
your strong base as you had

206
00:12:14,730 --> 00:12:18,370
weak acid, then you'll have the
conjugate base around, and

207
00:12:18,370 --> 00:12:21,320
the p h will be greater
than 7.

208
00:12:21,320 --> 00:12:24,750
So in working the problems, if
you get an answer with this

209
00:12:24,750 --> 00:12:27,620
type of titration problem that's
different than that for

210
00:12:27,620 --> 00:12:30,070
p h at the equivalence point,
you're going to know that you

211
00:12:30,070 --> 00:12:31,590
did something wrong,
you need to go back

212
00:12:31,590 --> 00:12:33,650
and check your math.

213
00:12:33,650 --> 00:12:37,530
Another big difference has to
do with the curve shape down

214
00:12:37,530 --> 00:12:40,760
here, and so you notice
a difference over

215
00:12:40,760 --> 00:12:45,410
here than over there.

216
00:12:45,410 --> 00:12:50,060
And in a titration that involves
a weak acid in a

217
00:12:50,060 --> 00:12:53,490
strong base, you have a part of
the curve that's known as a

218
00:12:53,490 --> 00:12:57,950
buffering region, and the p
h is fairly flat in this

219
00:12:57,950 --> 00:12:59,930
buffering region as
shown down here.

220
00:12:59,930 --> 00:13:02,520
So that's in contrast, there's
no such buffering

221
00:13:02,520 --> 00:13:04,230
region on this side.

222
00:13:04,230 --> 00:13:06,580
So here the p h will go
up, it'll level off,

223
00:13:06,580 --> 00:13:07,900
and then go up again.

224
00:13:07,900 --> 00:13:11,240
And this, for some of you, is
probably the frustration in

225
00:13:11,240 --> 00:13:14,510
doing acid based titrations in
lab, because you're adding and

226
00:13:14,510 --> 00:13:16,350
nothing's happening and
nothing's happening and

227
00:13:16,350 --> 00:13:18,700
nothing's happening, and you're
in this region, then

228
00:13:18,700 --> 00:13:20,090
all of a sudden you add
just a little more

229
00:13:20,090 --> 00:13:21,740
and you're up here.

230
00:13:21,740 --> 00:13:23,930
So notice how steep
that is over here.

231
00:13:23,930 --> 00:13:26,940
So sometimes when you're in the
buffering region, it seems

232
00:13:26,940 --> 00:13:29,170
like you're never going to reach
the end of the titration

233
00:13:29,170 --> 00:13:32,480
and then it'll happen
all too quickly.

234
00:13:32,480 --> 00:13:35,690
So buffering region, remember a
buffer is something that has

235
00:13:35,690 --> 00:13:39,980
a conjugate, weak acid and weak
base pair, and then in a

236
00:13:39,980 --> 00:13:43,730
buffering region, the p h
pretty much stays fairly

237
00:13:43,730 --> 00:13:45,530
constant in that region.

238
00:13:45,530 --> 00:13:48,840
It acts as a buffer,
neutralizing the p h,

239
00:13:48,840 --> 00:13:52,670
maintaining the p h by being a
source or sink of protons, and

240
00:13:52,670 --> 00:13:56,010
so here the p h then is staying
constant in that

241
00:13:56,010 --> 00:13:58,360
buffering region.

242
00:13:58,360 --> 00:13:59,720
So those are some of
the differences

243
00:13:59,720 --> 00:14:02,710
between the type of curves.

244
00:14:02,710 --> 00:14:06,230
Another point that I will
mention or term I will mention

245
00:14:06,230 --> 00:14:09,805
that has to do with weak acid in
strong base or a weak base

246
00:14:09,805 --> 00:14:14,060
in strong acid is this 1/2
equivalence point concept.

247
00:14:14,060 --> 00:14:17,370
So 1/2 equivalence point you've
added 1/2 of the amount

248
00:14:17,370 --> 00:14:20,030
of strong base that you need
to get to the equivalence

249
00:14:20,030 --> 00:14:21,940
point, and that's right
in the middle of

250
00:14:21,940 --> 00:14:23,290
that buffering region.

251
00:14:23,290 --> 00:14:25,090
So that's another point where
you'll be asked to

252
00:14:25,090 --> 00:14:28,410
calculate the p h.

253
00:14:28,410 --> 00:14:34,970
So now let's look at different
points in a titration.

254
00:14:34,970 --> 00:14:37,790
So, first let's walk through
and just think

255
00:14:37,790 --> 00:14:40,340
about what is happening.

256
00:14:40,340 --> 00:14:43,710
So when we start in this
titration of a weak acid in a

257
00:14:43,710 --> 00:14:48,540
strong base, before we've added
any of the strong base,

258
00:14:48,540 --> 00:14:50,370
all we have is a weak acid.

259
00:14:50,370 --> 00:14:53,640
So it is a weak acid in
water type problem.

260
00:14:53,640 --> 00:14:57,780
And so here I've drawn our acid,
and the acid has its

261
00:14:57,780 --> 00:15:00,470
proton, which is going
to give up when you

262
00:15:00,470 --> 00:15:03,310
start doing the titration.

263
00:15:03,310 --> 00:15:06,790
So that's what we have
at zero volume.

264
00:15:06,790 --> 00:15:10,180
Then we start adding our strong
base, and the strong

265
00:15:10,180 --> 00:15:13,230
base is going to react with
the acid, one-to-one

266
00:15:13,230 --> 00:15:15,490
stoichiometry, it's
a strong base.

267
00:15:15,490 --> 00:15:20,710
It'll pull off protons off the
same number of moles of the

268
00:15:20,710 --> 00:15:23,580
strong acid as the number of
moles of the strong base that

269
00:15:23,580 --> 00:15:24,540
were added.

270
00:15:24,540 --> 00:15:28,010
And so then, you'll start to
have a mixture of your

271
00:15:28,010 --> 00:15:31,490
conjugates, your weak acid
and your conjugate base.

272
00:15:31,490 --> 00:15:33,310
So the base is a minus here.

273
00:15:33,310 --> 00:15:35,950
And so if you have a mixture
of a weak acid in its

274
00:15:35,950 --> 00:15:38,700
conjugate base, that's a buffer,
and so you'll move in

275
00:15:38,700 --> 00:15:40,810
to the buffering region here.

276
00:15:40,810 --> 00:15:44,520
So that's at any volume that is
greater than zero and less

277
00:15:44,520 --> 00:15:47,570
than the equivalence point
is going to be

278
00:15:47,570 --> 00:15:52,290
around in that region.

279
00:15:52,290 --> 00:15:54,990
Then we have a special category
of the buffering

280
00:15:54,990 --> 00:15:57,820
region, which is when you've
added the volume to get to the

281
00:15:57,820 --> 00:15:59,670
1/2 equivalence point.

282
00:15:59,670 --> 00:16:03,070
And when you've done that, you
will have converted 1/2 of the

283
00:16:03,070 --> 00:16:06,840
weak acid it its conjugate base,
so you'll have equal

284
00:16:06,840 --> 00:16:10,680
number of moles of your weak
acid as moles of the

285
00:16:10,680 --> 00:16:11,210
conjugate base --

286
00:16:11,210 --> 00:16:14,350
1/2 has been converted.

287
00:16:14,350 --> 00:16:18,490
And so that's a special
category right there.

288
00:16:18,490 --> 00:16:20,320
Then you get to the
equivalence point.

289
00:16:20,320 --> 00:16:22,600
At the equivalence point, you've
added the same number

290
00:16:22,600 --> 00:16:26,000
of moles of strong base as the
number of moles of weak acid

291
00:16:26,000 --> 00:16:30,420
you have, so you've converted
all of your weak acid to it's

292
00:16:30,420 --> 00:16:31,500
conjugate base.

293
00:16:31,500 --> 00:16:34,730
So all you have is conjugate
base now, and so that's

294
00:16:34,730 --> 00:16:38,440
controlling the p h, so the p
h should be greater than 7.

295
00:16:38,440 --> 00:16:42,990
So that's a weak base
in water problem.

296
00:16:42,990 --> 00:16:48,320
And if you keep going, then
you're going to end up with a

297
00:16:48,320 --> 00:16:50,830
strong base in water problem.

298
00:16:50,830 --> 00:16:55,280
The weak base will still be
around, but it will be

299
00:16:55,280 --> 00:16:58,020
negligibly affecting the p h
compared to the fact that

300
00:16:58,020 --> 00:17:00,600
you're dumping strong acid
into your titration.

301
00:17:00,600 --> 00:17:03,870
And so that's this part
of the curve.

302
00:17:03,870 --> 00:17:08,590
So you see that in one type
of problem, one titration

303
00:17:08,590 --> 00:17:13,100
problem, you actually have a
lot of sub problems, or sub

304
00:17:13,100 --> 00:17:16,650
types of problems, you'll have
weak acid buffer, special

305
00:17:16,650 --> 00:17:20,430
category of buffer, a conjugate
base or a salt

306
00:17:20,430 --> 00:17:22,110
issue, and then a strong base.

307
00:17:22,110 --> 00:17:24,580
And this is one of the things
that people have trouble with

308
00:17:24,580 --> 00:17:27,530
in the titrations, because we
may not ask you to do all the

309
00:17:27,530 --> 00:17:30,270
points, we may just sort of
jump in somewhere, and say

310
00:17:30,270 --> 00:17:32,910
okay, what is the p h at the
equivalenced point, and you

311
00:17:32,910 --> 00:17:35,350
need to think about what's
happened to get to the

312
00:17:35,350 --> 00:17:36,620
equivalence point.

313
00:17:36,620 --> 00:17:39,550
Or we may jump in and ask you
about a region that would be

314
00:17:39,550 --> 00:17:42,020
in the buffering region, and you
have to remember that at

315
00:17:42,020 --> 00:17:44,850
that point you should have some
of the weak acid and also

316
00:17:44,850 --> 00:17:47,250
some of the conjugate
bases being formed.

317
00:17:47,250 --> 00:17:50,600
So, it seems like there are a
lot of different things, but

318
00:17:50,600 --> 00:17:53,720
there are only five types of
problems. But in a titration

319
00:17:53,720 --> 00:17:57,850
curve, you run into a lot of
those different types at

320
00:17:57,850 --> 00:18:02,480
different points
in the problem.

321
00:18:02,480 --> 00:18:04,730
So now let's go the other
direction and consider

322
00:18:04,730 --> 00:18:08,360
titration of a weak base
with a strong acid.

323
00:18:08,360 --> 00:18:10,100
So here's what that curve
would look like.

324
00:18:10,100 --> 00:18:12,440
You're going to start basic,
of course, because you're

325
00:18:12,440 --> 00:18:14,870
starting with a weak base,
you haven't added

326
00:18:14,870 --> 00:18:16,430
any strong acid yet.

327
00:18:16,430 --> 00:18:20,460
As you add strong acid,
the p h will decrease.

328
00:18:20,460 --> 00:18:23,450
Because it is a weak base, you
will be forming some of its

329
00:18:23,450 --> 00:18:26,970
conjugate as you add the strong
acid, and so you'll go

330
00:18:26,970 --> 00:18:29,770
through a buffering region again
where the curve would be

331
00:18:29,770 --> 00:18:32,300
flat, where the p h will be
pretty much the same for

332
00:18:32,300 --> 00:18:33,790
region of time.

333
00:18:33,790 --> 00:18:37,010
Then the curve will drop again
and you'll get to the

334
00:18:37,010 --> 00:18:38,470
equivalence point.

335
00:18:38,470 --> 00:18:42,460
At the equivalence point, you've
added the same amount

336
00:18:42,460 --> 00:18:45,980
of moles of strong acid as you
had weak base, so all of your

337
00:18:45,980 --> 00:18:49,760
weak base is converted to its
conjugate acid, and so you

338
00:18:49,760 --> 00:18:53,580
should be acidic at the
equivalence point, and then

339
00:18:53,580 --> 00:18:55,980
the curve goes down.

340
00:18:55,980 --> 00:18:57,870
So again, we can think
about this in

341
00:18:57,870 --> 00:18:59,750
terms of what is happening.

342
00:18:59,750 --> 00:19:03,030
In the beginning it's just a
weak base in water problem,

343
00:19:03,030 --> 00:19:07,270
but as you add strong acid,
you were pronating some of

344
00:19:07,270 --> 00:19:10,560
your base and forming its
conjugate acid here, and

345
00:19:10,560 --> 00:19:13,170
you're in the going to be
in the buffering region.

346
00:19:13,170 --> 00:19:17,310
Then at the 1/2 equivalence
point, you've added enough

347
00:19:17,310 --> 00:19:20,980
moles of strong acid to convert
1/2 of the weak base

348
00:19:20,980 --> 00:19:24,250
to its conjugate, so those are
going to be equal to each

349
00:19:24,250 --> 00:19:26,900
other -- the number of moles
of the weak base and the

350
00:19:26,900 --> 00:19:29,210
number of moles of its
conjugate acid.

351
00:19:29,210 --> 00:19:32,410
At the equivalence point, you've
converted all of the

352
00:19:32,410 --> 00:19:35,840
weak base you started with to
its conjugate acid, so it'll

353
00:19:35,840 --> 00:19:39,410
be a weak acid in water problem,
and then at the end

354
00:19:39,410 --> 00:19:41,950
it's strong acid.

355
00:19:41,950 --> 00:19:45,400
So the trick is to recognizing
what type of problem you're

356
00:19:45,400 --> 00:19:48,590
being asked to do, and a lot
of times if people get a

357
00:19:48,590 --> 00:19:51,570
question and they just write
down OK, at this point in the

358
00:19:51,570 --> 00:19:53,790
titration curve, it's
going to be a weak

359
00:19:53,790 --> 00:19:55,130
base in water problem.

360
00:19:55,130 --> 00:19:58,620
And just writing that down, most
of the time if you get

361
00:19:58,620 --> 00:20:01,570
that far, you do the rest of
the problem correctly.

362
00:20:01,570 --> 00:20:05,950
So just identifying the type,
there are only 5, of problems

363
00:20:05,950 --> 00:20:12,040
gets you a long way to getting
the right answer.

364
00:20:12,040 --> 00:20:16,150
So let's do an example.

365
00:20:16,150 --> 00:20:19,470
We're going to titrate a weak
acid with a strong base.

366
00:20:19,470 --> 00:20:25,260
We have 25 mils of 0.1 molar
acid with 0.15 moles of a

367
00:20:25,260 --> 00:20:33,040
strong base, n a o h, we're
given the k a for the acid.

368
00:20:33,040 --> 00:20:39,190
First we start with 0 mils
of the strong base added.

369
00:20:39,190 --> 00:20:45,240
So what type of problem
is this?

370
00:20:45,240 --> 00:20:48,140
It's a weak acid problem.

371
00:20:48,140 --> 00:20:51,330
So we know how to write the
equation for a weak acid or

372
00:20:51,330 --> 00:20:52,830
for an acid in water.

373
00:20:52,830 --> 00:20:57,940
We have the acid in water going
to hydronium ions and a

374
00:20:57,940 --> 00:21:01,710
conjugate base.

375
00:21:01,710 --> 00:21:03,900
So weak acid.

376
00:21:03,900 --> 00:21:10,415
For weak acid, we're going to
use our k a, and we're going

377
00:21:10,415 --> 00:21:13,770
to set up our equilibrium
expression.

378
00:21:13,770 --> 00:21:18,720
So here we have 0.1
molar of our acid.

379
00:21:18,720 --> 00:21:21,170
We're going to have some
of that go away in the

380
00:21:21,170 --> 00:21:26,080
equilibrium, forming hydronium
ion and some conjugate base,

381
00:21:26,080 --> 00:21:30,160
and so we know we have
expressions for the

382
00:21:30,160 --> 00:21:32,550
concentrations at equilibrium.

383
00:21:32,550 --> 00:21:36,930
And we can use our k a, k a
for acid, it's a weak acid

384
00:21:36,930 --> 00:21:42,150
problem, and we can look at
products over reactants.

385
00:21:42,150 --> 00:21:44,850
So, see, now we're doing a
titration problem, but you

386
00:21:44,850 --> 00:21:47,330
already know how to do this
problem because we've seen a

387
00:21:47,330 --> 00:21:50,140
weak acid in water
problem before.

388
00:21:50,140 --> 00:21:56,710
So we have x squared over
0.10 minus x here.

389
00:21:56,710 --> 00:22:03,070
We can assume x is small, and
get rid of this minus x, and

390
00:22:03,070 --> 00:22:05,770
then later go back and check
it, so that just makes the

391
00:22:05,770 --> 00:22:07,790
math a little bit easier.

392
00:22:07,790 --> 00:22:13,900
And we can solve for x and then
we can check -- we can

393
00:22:13,900 --> 00:22:21,330
take this value, 0.00421 over
0.1 and see whether that's

394
00:22:21,330 --> 00:22:24,500
less than 5%, it's
close but it is.

395
00:22:24,500 --> 00:22:26,930
So that assumption is OK.

396
00:22:26,930 --> 00:22:30,240
If it wasn't, what would
we have to do?

397
00:22:30,240 --> 00:22:32,800
Quadratic equation.

398
00:22:32,800 --> 00:22:35,860
All right, so now, here's
a sig fig question.

399
00:22:35,860 --> 00:23:20,540
Tell me how many sig figs
this p h actually has.

400
00:23:20,540 --> 00:23:42,580
OK, 10 seconds.

401
00:23:42,580 --> 00:23:46,460
So, in the first part of the
problem we had a concentration

402
00:23:46,460 --> 00:23:51,140
that had 2 significant figures,
the 0.10 molar.

403
00:23:51,140 --> 00:23:54,260
Sometimes later, people have
extra significant figures that

404
00:23:54,260 --> 00:23:57,380
they're carrying along, but we
had those 2, and so we're

405
00:23:57,380 --> 00:24:01,630
going to have 2 after the
decimal point then in the

406
00:24:01,630 --> 00:24:05,800
answer of the p h.

407
00:24:05,800 --> 00:24:07,990
So again, the number of
significant figures that are

408
00:24:07,990 --> 00:24:09,730
limiting are going
to be the number

409
00:24:09,730 --> 00:24:15,650
after the decimal point.

410
00:24:15,650 --> 00:24:20,460
All right, so we have one
p h value, and now we're

411
00:24:20,460 --> 00:24:21,120
going to move on.

412
00:24:21,120 --> 00:24:26,360
So let me just put our
one p h value down.

413
00:24:26,360 --> 00:24:37,720
We have volume of strong base,
and p h over here, and we're

414
00:24:37,720 --> 00:24:40,760
starting here with
zero moles added.

415
00:24:40,760 --> 00:24:43,020
We have a p h of 2 .

416
00:24:43,020 --> 00:24:43,920
38.

417
00:24:43,920 --> 00:24:49,990
It's a weak acid, so it should
be an acidic p h, which it is.

418
00:24:49,990 --> 00:24:55,220
All right, so now let's move
into the titration problem,

419
00:24:55,220 --> 00:24:56,510
and now 5 .

420
00:24:56,510 --> 00:25:02,050
0 mils of the strong base have
been added, and we need to

421
00:25:02,050 --> 00:25:04,790
find what the p h is now.

422
00:25:04,790 --> 00:25:09,070
So it's a strong base, so it's
going to react almost

423
00:25:09,070 --> 00:25:10,980
completely, that's
our assumption.

424
00:25:10,980 --> 00:25:13,320
If it's strong, it
goes completely.

425
00:25:13,320 --> 00:25:18,120
And so, the number of moles of
the strong base that we add

426
00:25:18,120 --> 00:25:22,790
will convert all of the same
number of moles of our acid

427
00:25:22,790 --> 00:25:25,770
over to its conjugate.

428
00:25:25,770 --> 00:25:29,320
So we can just do a
subtraction then.

429
00:25:29,320 --> 00:25:31,810
So first, we need to know
the initial moles of the

430
00:25:31,810 --> 00:25:33,160
acid that we had.

431
00:25:33,160 --> 00:25:36,580
We had 25 mils, 0.10 molar.

432
00:25:36,580 --> 00:25:40,520
We calculate the number of moles
for the hydroxide added,

433
00:25:40,520 --> 00:25:44,800
we added 5 mils, it was 0.15
molar, and so we can calculate

434
00:25:44,800 --> 00:25:48,280
the number of moles of the
strong base that were added.

435
00:25:48,280 --> 00:25:52,590
So the strong base will react
completely with the same

436
00:25:52,590 --> 00:25:55,470
number of moles of
the weak acid.

437
00:25:55,470 --> 00:25:58,470
And we're going to do then -- we
have the moles of the weak

438
00:25:58,470 --> 00:26:01,710
acid here, minus the number of
moles of the strong base we've

439
00:26:01,710 --> 00:26:03,860
added, and so we're
going to have 1 .

440
00:26:03,860 --> 00:26:09,180
75 times 10 to the minus 3 moles
of the weak acid left.

441
00:26:09,180 --> 00:26:14,940
So, then how many moles of the
conjugate base will be formed

442
00:26:14,940 --> 00:26:19,610
by this reaction?

443
00:26:19,610 --> 00:26:28,920
What do you think?

444
00:26:28,920 --> 00:26:29,780
Same number.

445
00:26:29,780 --> 00:26:39,530
So 0.75 times 10
to the minus 3.

446
00:26:39,530 --> 00:26:42,930
So always remember that in
these titration problems,

447
00:26:42,930 --> 00:26:46,530
nothing has been added yet,
you're at zero mils added.

448
00:26:46,530 --> 00:26:48,780
Some amount of some subtractions
are going to have

449
00:26:48,780 --> 00:26:50,680
to occur because something
has happened.

450
00:26:50,680 --> 00:26:54,000
You've converted something,
things are different than when

451
00:26:54,000 --> 00:26:54,660
you started.

452
00:26:54,660 --> 00:27:01,030
All right, so now we have weak
acid and we have moles of its

453
00:27:01,030 --> 00:27:07,090
conjugate, what type
of problem is this?

454
00:27:07,090 --> 00:27:14,380
If you have a weak acid and
its conjugate base --

455
00:27:14,380 --> 00:27:16,600
buffer, right.

456
00:27:16,600 --> 00:27:19,790
So we're going to do a buffer
problem and we need to know

457
00:27:19,790 --> 00:27:25,440
the molarity first. So we have
moles over volume -- again,

458
00:27:25,440 --> 00:27:29,610
the volume, you had 25 mils to
begin with, you added 5 more.

459
00:27:29,610 --> 00:27:33,080
So you have to have the total
volume 30 mils, and we can

460
00:27:33,080 --> 00:27:37,100
calculate then the
concentrations of both.

461
00:27:37,100 --> 00:27:40,870
Now we can set up our
equilibrium table, and this

462
00:27:40,870 --> 00:27:44,130
looks like a buffer problem
because it is, and by looking

463
00:27:44,130 --> 00:27:46,556
like a buffer problem you
something over here, you have

464
00:27:46,556 --> 00:27:49,560
your weak acid over here, but
you have something over here

465
00:27:49,560 --> 00:27:52,140
now, it's not zero now,
we're starting with

466
00:27:52,140 --> 00:27:53,900
some conjugate base.

467
00:27:53,900 --> 00:28:01,110
So we have 0.0583 minus x on one
side, and we 0.025 molar

468
00:28:01,110 --> 00:28:05,240
plus x on the other side.

469
00:28:05,240 --> 00:28:07,250
We can use k a again.

470
00:28:07,250 --> 00:28:12,150
This is set up as an acid in
water going to hydronium ions

471
00:28:12,150 --> 00:28:18,470
and conjugate base, so we can
use our k a, set things up,

472
00:28:18,470 --> 00:28:22,460
and we can always say let's
see if x is small, make an

473
00:28:22,460 --> 00:28:24,890
assumption, check it later.

474
00:28:24,890 --> 00:28:27,860
That'll simplify the math.

475
00:28:27,860 --> 00:28:30,320
So we get rid of the plus
x and the minus x.

476
00:28:30,320 --> 00:28:33,300
Again, we're saying that if
x is small, the initial

477
00:28:33,300 --> 00:28:35,930
concentrations are going to be
more or less the same as the

478
00:28:35,930 --> 00:28:40,540
concentrations after the
equilibration occurs.

479
00:28:40,540 --> 00:28:42,670
And we can calculate 4 .

480
00:28:42,670 --> 00:28:46,540
13 times 10 to the minus
4, as x, that is a

481
00:28:46,540 --> 00:28:48,100
pretty small number.

482
00:28:48,100 --> 00:28:51,460
And we have to check it, and
yup, it's small enough, it's

483
00:28:51,460 --> 00:28:56,730
under 5%, so that's OK.

484
00:28:56,730 --> 00:28:58,080
So now we can plug this in.

485
00:28:58,080 --> 00:29:01,380
X is our hydronium ion
concentration minus log of the

486
00:29:01,380 --> 00:29:05,140
hydronium ion concentration
is p h, and we can

487
00:29:05,140 --> 00:29:07,680
calculate p h to 3 .

488
00:29:07,680 --> 00:29:12,880
38 -- again, we're limited by
two significant figures in the

489
00:29:12,880 --> 00:29:14,410
concentration.

490
00:29:14,410 --> 00:29:19,710
So now we've added 5 mils down
here, and our p h has gone up

491
00:29:19,710 --> 00:29:24,090
a little bit, it's now at 3 .

492
00:29:24,090 --> 00:29:30,080
38 over here.

493
00:29:30,080 --> 00:29:33,510
There's another option
for a buffer problem.

494
00:29:33,510 --> 00:29:39,390
What's the one equation
in this unit?

495
00:29:39,390 --> 00:29:43,770
Our friend Henderson
Hasselbalch.

496
00:29:43,770 --> 00:29:47,300
And yes, you can use that here
too, assuming that you check

497
00:29:47,300 --> 00:29:49,570
the assumption and it's OK.

498
00:29:49,570 --> 00:29:51,670
Most people will prefer
to do this

499
00:29:51,670 --> 00:29:54,120
because it is a bit easier.

500
00:29:54,120 --> 00:29:58,290
So, you weren't given, though,
the p k a in this problem, you

501
00:29:58,290 --> 00:30:03,110
were given the k a, so pretty
easy to calculate -- minus log

502
00:30:03,110 --> 00:30:05,500
of the k a is the p k a.

503
00:30:05,500 --> 00:30:08,210
So you can calculate
that, put that in.

504
00:30:08,210 --> 00:30:10,320
You have your concentrations
and it should be

505
00:30:10,320 --> 00:30:13,430
concentrations, but you may
notice that if you actually

506
00:30:13,430 --> 00:30:16,460
had moles the volume
would cancel here.

507
00:30:16,460 --> 00:30:19,760
So here are the concentrations,
but with the

508
00:30:19,760 --> 00:30:22,600
same volume, the volume
term does cancel.

509
00:30:22,600 --> 00:30:25,680
It makes this a little faster
and it gives the same answer,

510
00:30:25,680 --> 00:30:27,370
which is great.

511
00:30:27,370 --> 00:30:30,800
To use Henderson Hasselbalch you
also need the 5% rule to

512
00:30:30,800 --> 00:30:33,270
be true, because Henderson
Hasselbalch is

513
00:30:33,270 --> 00:30:35,100
assuming that x is small.

514
00:30:35,100 --> 00:30:37,770
It's assuming that the initial
concentrations and the

515
00:30:37,770 --> 00:30:40,320
concentrations after
equilibrium

516
00:30:40,320 --> 00:30:42,180
are about the same.

517
00:30:42,180 --> 00:30:44,280
So we can check the
assumption.

518
00:30:44,280 --> 00:30:47,460
We can back-calculate the
hydronium ion concentration,

519
00:30:47,460 --> 00:30:50,270
which would be x, and see if
it's small, we already know it

520
00:30:50,270 --> 00:30:52,080
is, so it's OK.

521
00:30:52,080 --> 00:30:56,090
So there are 2 options for
buffer problems, but do not

522
00:30:56,090 --> 00:30:58,850
use the Henderson Hasselbalch
equation when it isn't in the

523
00:30:58,850 --> 00:31:02,960
buffering region, it
doesn't hold then.

524
00:31:02,960 --> 00:31:04,990
So again, you check the
assumption, and if

525
00:31:04,990 --> 00:31:06,100
it's OK, it's fine.

526
00:31:06,100 --> 00:31:09,970
If not, you need to use option
one and you need to use the

527
00:31:09,970 --> 00:31:14,730
quadratic equation.

528
00:31:14,730 --> 00:31:18,490
All right, so buffering
region.

529
00:31:18,490 --> 00:31:22,190
Now we're at the special kind
of problem in the buffering

530
00:31:22,190 --> 00:31:25,470
region, the 1/2 equivalence
point.

531
00:31:25,470 --> 00:31:30,260
So here you've added 1/2 the
number of moles of the strong

532
00:31:30,260 --> 00:31:32,840
base to convert 1/2
the moles of the

533
00:31:32,840 --> 00:31:35,850
weak acid to its conjugate.

534
00:31:35,850 --> 00:31:40,330
So at this point, the
concentration of h a equals

535
00:31:40,330 --> 00:31:44,680
the concentration of a minus
-- equal number of moles in

536
00:31:44,680 --> 00:31:47,460
the same volume, those
are equal.

537
00:31:47,460 --> 00:31:51,140
You can use Henderson
Hasselbalch here, and find

538
00:31:51,140 --> 00:31:55,700
that if they're equal, you're
talking about minus log of 1,

539
00:31:55,700 --> 00:32:02,390
so the p h is going to
equal the p k a.

540
00:32:02,390 --> 00:32:05,610
And you're done with this
type of problem.

541
00:32:05,610 --> 00:32:09,560
I have been known to put 1/2
equivalence problems on an

542
00:32:09,560 --> 00:32:12,900
exam, because exams are often
long, you have only 50

543
00:32:12,900 --> 00:32:15,090
minutes, there's lots of
different type of problems,

544
00:32:15,090 --> 00:32:17,865
and this problem should
not take you a

545
00:32:17,865 --> 00:32:19,070
long amount of time.

546
00:32:19,070 --> 00:32:22,870
You do not have to prove to
me that this is true.

547
00:32:22,870 --> 00:32:26,750
All you need to remember, 1/2
equivalence point, p h equals

548
00:32:26,750 --> 00:32:32,100
p k a, and if you calculate
the p k a, you're done.

549
00:32:32,100 --> 00:32:35,160
So this is a short
type of problem.

550
00:32:35,160 --> 00:32:38,540
If you remember the definition
of 1/2 equivalence point, it's

551
00:32:38,540 --> 00:32:40,150
easy to do.

552
00:32:40,150 --> 00:32:45,480
So now we have another
number, so 3 .

553
00:32:45,480 --> 00:32:58,290
75, and we're working
on our curve.

554
00:32:58,290 --> 00:33:01,950
Now let's move to the
equivalence point.

555
00:33:01,950 --> 00:33:05,780
At the equivalence point, you've
added the same number

556
00:33:05,780 --> 00:33:09,270
of moles of your strong base
as you had weak acid.

557
00:33:09,270 --> 00:33:13,880
So you've converted all of
your weak acid to its

558
00:33:13,880 --> 00:33:18,020
conjugate base.

559
00:33:18,020 --> 00:33:20,240
So the p h should be
greater than 7.

560
00:33:20,240 --> 00:33:24,220
Now all you have is conjugate
base, that's basic, p h should

561
00:33:24,220 --> 00:33:28,920
be greater than 7.

562
00:33:28,920 --> 00:33:33,390
So when you are doing this
titration, you have your weak

563
00:33:33,390 --> 00:33:35,710
acid and your strong base.

564
00:33:35,710 --> 00:33:40,950
You're going to be forming a
salt here, and a salt problem,

565
00:33:40,950 --> 00:33:43,980
you can tell me about salts.

566
00:33:43,980 --> 00:33:49,320
And so, just remind me, what
does the n a plus contribute

567
00:33:49,320 --> 00:33:54,200
to the p h here.

568
00:33:54,200 --> 00:33:55,940
It's going to be neutral.

569
00:33:55,940 --> 00:34:00,540
And what about this
guy down here?

570
00:34:00,540 --> 00:34:02,450
Yeah, so it's going
to be basic.

571
00:34:02,450 --> 00:34:06,670
So, the sodium, anything group
1, group 2, no effect on p h,

572
00:34:06,670 --> 00:34:07,670
they're neutral.

573
00:34:07,670 --> 00:34:10,880
But if you have a conjugate base
of a weak acid, that's

574
00:34:10,880 --> 00:34:13,510
going to be basic.

575
00:34:13,510 --> 00:34:17,640
Salt problems, really just
part of what you

576
00:34:17,640 --> 00:34:22,710
already know about.

577
00:34:22,710 --> 00:34:24,010
So always check your work.

578
00:34:24,010 --> 00:34:27,350
If your p h doesn't make sense
from what you know, you might

579
00:34:27,350 --> 00:34:31,980
have made a math mistake.

580
00:34:31,980 --> 00:34:35,500
So let's calculate the actual p
h at the equivalence point.

581
00:34:35,500 --> 00:34:37,940
We know that it should
be basic, but what

582
00:34:37,940 --> 00:34:41,590
is it going to be?

583
00:34:41,590 --> 00:34:45,880
So first, we need to know how
much of the strong base we had

584
00:34:45,880 --> 00:34:50,460
to add, because we need to
know about all the moles.

585
00:34:50,460 --> 00:34:53,110
So how much of this did
we need to add.

586
00:34:53,110 --> 00:34:56,040
So we needed to add enough of
the strong base that you

587
00:34:56,040 --> 00:34:57,230
converted all of the
moles of the

588
00:34:57,230 --> 00:34:59,540
weak acid to its conjugate.

589
00:34:59,540 --> 00:35:00,650
So we had 2 .

590
00:35:00,650 --> 00:35:04,280
5 times 10 to the minus 3
moles of our weak acid.

591
00:35:04,280 --> 00:35:07,810
So that's all going to be
converted to the moles of the

592
00:35:07,810 --> 00:35:10,430
conjugate base, and so that's
going to be equal to the

593
00:35:10,430 --> 00:35:12,470
number of moles we
needed to do it.

594
00:35:12,470 --> 00:35:13,900
So we needed 2 .

595
00:35:13,900 --> 00:35:17,610
5 times 10 to the minus 3 moles
of our strong base to do

596
00:35:17,610 --> 00:35:19,190
that complete conversion.

597
00:35:19,190 --> 00:35:22,260
We know the concentration
of the base was 0.15 .

598
00:35:22,260 --> 00:35:23,650
So we would have needed 1 .

599
00:35:23,650 --> 00:35:28,880
67 times 10 to the minus 2
liters of this concentration

600
00:35:28,880 --> 00:35:32,520
added to reach the equivalence
point.

601
00:35:32,520 --> 00:35:34,890
So then the total volume that
we're going to have at the

602
00:35:34,890 --> 00:35:38,020
equivalence point is the 25
mils that we had to begin

603
00:35:38,020 --> 00:35:40,750
with, plus this 16 .

604
00:35:40,750 --> 00:35:45,850
7 mils to make this final,
total volume.

605
00:35:45,850 --> 00:35:48,700
And remember, you always need
to think, what is the total

606
00:35:48,700 --> 00:35:51,340
volume, how much has been added
to get to this point in

607
00:35:51,340 --> 00:35:54,590
the titration curve.

608
00:35:54,590 --> 00:35:57,770
Then we can calculate molarity,
so we know how many

609
00:35:57,770 --> 00:36:00,850
moles of conjugate base have
been formed, and we know the

610
00:36:00,850 --> 00:36:05,000
new volume, so we can calculate
the concentration of

611
00:36:05,000 --> 00:36:08,500
the conjugate base.

612
00:36:08,500 --> 00:36:12,970
So now, you can help me
solve this problem.

613
00:36:12,970 --> 00:36:43,520
Set up an equation for
me to solve it.

614
00:36:43,520 --> 00:37:02,280
Let's take 10 seconds.

615
00:37:02,280 --> 00:37:05,520
That's the best score
we've had today.

616
00:37:05,520 --> 00:37:06,540
Yup.

617
00:37:06,540 --> 00:37:11,010
So now we're talking about
a conjugate base.

618
00:37:11,010 --> 00:37:16,920
So we have converted all of the
weak acid to the conjugate

619
00:37:16,920 --> 00:37:23,190
base, and so it's a weak base
in water problem, so we're

620
00:37:23,190 --> 00:37:26,060
going to talk about a k b.

621
00:37:26,060 --> 00:37:29,810
If you were only given the k a
for this problem, how would

622
00:37:29,810 --> 00:37:35,170
you find k b -- what
interconnects k a and k b?

623
00:37:35,170 --> 00:37:37,070
K w, right.

624
00:37:37,070 --> 00:37:39,610
So you can calculate, here it's
given to you, but you

625
00:37:39,610 --> 00:37:43,040
could calculate it if you had
a calculator, and you would

626
00:37:43,040 --> 00:37:44,670
find that this is true.

627
00:37:44,670 --> 00:37:46,980
Now it's a weak base
in water problem.

628
00:37:46,980 --> 00:37:48,730
We're not in the buffering
region anymore.

629
00:37:48,730 --> 00:37:51,660
We've converted all of our weak
acid to the conjugate.

630
00:37:51,660 --> 00:37:54,670
So it's a weak base
in water problem.

631
00:37:54,670 --> 00:37:59,200
So we have x squared, 0.06, that
was the concentration we

632
00:37:59,200 --> 00:38:04,750
calculated, minus x.

633
00:38:04,750 --> 00:38:08,520
So again, think about what
type of problem it is.

634
00:38:08,520 --> 00:38:12,710
So again, weak base in water
problem -- x squared

635
00:38:12,710 --> 00:38:16,720
over 0.06 minus x.

636
00:38:16,720 --> 00:38:22,560
And we can assume that x is
small, and calculate a value

637
00:38:22,560 --> 00:38:27,380
for x, which is 0.83 times 10 to
the minus 6, and then we're

638
00:38:27,380 --> 00:38:31,330
going to calculate p o h,
because now x is the hydroxide

639
00:38:31,330 --> 00:38:33,320
ion concentration.

640
00:38:33,320 --> 00:38:38,750
Because in a weak base in water
problem, here in this

641
00:38:38,750 --> 00:38:42,240
type of problem, the base, and
here is your acid -- the

642
00:38:42,240 --> 00:38:45,620
conjugate of this acid is the
base, hydroxide, and the

643
00:38:45,620 --> 00:38:50,010
conjugate of this weak base is
its conjugate acid over here,

644
00:38:50,010 --> 00:38:54,020
so now when we are solving for
x, we're solving for hydroxide

645
00:38:54,020 --> 00:38:58,510
ion in concentration, so we're
calculating a p o h, which

646
00:38:58,510 --> 00:39:02,800
then we can calculate
a p h from.

647
00:39:02,800 --> 00:39:05,510
So we can take 14 minus 5 .

648
00:39:05,510 --> 00:39:08,060
74 and get our value.

649
00:39:08,060 --> 00:39:12,830
And it's bigger than neutral,
it's 8, it's basic, and that

650
00:39:12,830 --> 00:39:17,380
makes sense, it is a weak
base in water problem.

651
00:39:17,380 --> 00:39:19,410
So, let's see, it's 8 .

652
00:39:19,410 --> 00:39:26,320
26, so now we're up here in our
curve, and we're at 8 .

653
00:39:26,320 --> 00:39:32,550
26, and that's going to be
greater than 7 for this type

654
00:39:32,550 --> 00:39:34,710
of problem.

655
00:39:34,710 --> 00:39:38,510
So that makes sense,
it's good.

656
00:39:38,510 --> 00:39:44,140
Greater than 7 is what
we want to see.

657
00:39:44,140 --> 00:39:48,800
So now, you've gone too far --
you've passed the equivalence

658
00:39:48,800 --> 00:39:55,990
point, and you keep adding
your strong base in.

659
00:39:55,990 --> 00:39:59,580
Now you still have some of the
weak conjugate base around.

660
00:39:59,580 --> 00:40:02,920
So you still have this around,
but you only have 1 .

661
00:40:02,920 --> 00:40:05,500
83 times 10 to the minus
6 molar of it.

662
00:40:05,500 --> 00:40:08,980
So very little amount
-- x is small.

663
00:40:08,980 --> 00:40:12,440
So your p h is going to be
dictated by the amount of

664
00:40:12,440 --> 00:40:18,170
extra strong base
you're adding.

665
00:40:18,170 --> 00:40:22,770
So this is similar, then, to a
strong acid or strong base in

666
00:40:22,770 --> 00:40:25,480
water problem.

667
00:40:25,480 --> 00:40:30,430
So if you're 5 mils past the
equivalence point, 5 mils

668
00:40:30,430 --> 00:40:33,430
times your concentration of
a strong base, so you

669
00:40:33,430 --> 00:40:34,980
have extra, 7 .

670
00:40:34,980 --> 00:40:38,710
5 times 10 to the minus
4 moles extra.

671
00:40:38,710 --> 00:40:42,480
So then you need to calculate
a concentration of that, and

672
00:40:42,480 --> 00:40:45,730
so you remember the whole volume
-- you're 5 mils past,

673
00:40:45,730 --> 00:40:49,290
you had 25 miles to start with,
and you had to add 16 .

674
00:40:49,290 --> 00:40:52,360
7 mils to get to the
equivalence point.

675
00:40:52,360 --> 00:40:54,990
And you have, that's your
total volume, you get a

676
00:40:54,990 --> 00:40:58,510
concentration, that's your
concentration of hydroxide, it

677
00:40:58,510 --> 00:41:01,250
reacts completely, you don't
have to do any equilibrium

678
00:41:01,250 --> 00:41:02,210
table here.

679
00:41:02,210 --> 00:41:04,910
It's going complete,
it's a strong base.

680
00:41:04,910 --> 00:41:09,710
You could try adding that value
of your other weak base

681
00:41:09,710 --> 00:41:13,880
to this, but remember, that's
times 10 to the minus 6, so

682
00:41:13,880 --> 00:41:15,340
it's not going to
be significant

683
00:41:15,340 --> 00:41:16,830
with significant figures.

684
00:41:16,830 --> 00:41:20,090
So you can just use this value
-- plug it in to p o h,

685
00:41:20,090 --> 00:41:23,570
calculate it, and then
calculate p h.

686
00:41:23,570 --> 00:41:32,570
And so now we're somewhere
up here at p h 12 .

687
00:41:32,570 --> 00:41:39,880
21, 5 mils past.
And there we've

688
00:41:39,880 --> 00:41:46,260
worked a titration problem.

689
00:41:46,260 --> 00:41:49,460
So let's review what we saw.

690
00:41:49,460 --> 00:41:52,540
In the beginning, zero mils of
the strong base, we have a

691
00:41:52,540 --> 00:41:54,660
weak acid in water problem.

692
00:41:54,660 --> 00:41:57,790
We moved into the buffering
region where we had our weak

693
00:41:57,790 --> 00:42:01,570
acid and the conjugate base
of that weak acid.

694
00:42:01,570 --> 00:42:04,000
At the equivalence point, we've
converted all of the

695
00:42:04,000 --> 00:42:06,070
weak acid to the conjugate
base, so

696
00:42:06,070 --> 00:42:07,680
it's a weak base problem.

697
00:42:07,680 --> 00:42:09,960
And then beyond the equivalence
point, it's a

698
00:42:09,960 --> 00:42:12,160
strong base problem.

699
00:42:12,160 --> 00:42:18,530
That's what we've just worked.

700
00:42:18,530 --> 00:42:20,810
So, we can check these
all off now.

701
00:42:20,810 --> 00:42:25,590
You know how to do all of these
types of problems. And

702
00:42:25,590 --> 00:42:28,870
there are not that many, you
just need to figure out where

703
00:42:28,870 --> 00:42:30,650
to apply what.

704
00:42:30,650 --> 00:42:33,330
And if you can do that, you're
all set, this unit will be

705
00:42:33,330 --> 00:42:38,110
easy for you, and you can go
through and make me very happy

706
00:42:38,110 --> 00:42:39,130
on the exam.

707
00:42:39,130 --> 00:42:41,680
There's nothing -- well, there
are few things in life as

708
00:42:41,680 --> 00:42:45,870
beautiful to me as a perfectly
worked titration problem.

709
00:42:45,870 --> 00:42:49,930
It really, it brings me joy, and
I've had people write on

710
00:42:49,930 --> 00:42:53,580
the exam sometimes, "I hope
that my solution to this

711
00:42:53,580 --> 00:42:57,840
brings you joy." And I will
often write, "Yes, it does,"

712
00:42:57,840 --> 00:42:59,460
and put a smiley face.

713
00:42:59,460 --> 00:43:02,680
Because it really is nice to see
these beautifully worked.

714
00:43:02,680 --> 00:43:08,850
I know, I'm a little nerdy and
geeky, but after yesterday,

715
00:43:08,850 --> 00:43:16,220
being smart and a nerd and
a geek is cool again.

716
00:43:16,220 --> 00:43:20,530
All right, so let me just tell
you where we're going.

717
00:43:20,530 --> 00:43:23,850
We have five more minutes, and
actually that's perfect,

718
00:43:23,850 --> 00:43:29,020
because I can get through some
rules in those 5 minutes.

719
00:43:29,020 --> 00:43:31,630
So let's do 5 minutes
of rules.

720
00:43:31,630 --> 00:43:35,830
Oxidation reduction doesn't have
a lot of rules, so five

721
00:43:35,830 --> 00:43:38,950
minutes is actually all
we need to do that.

722
00:43:38,950 --> 00:43:42,030
Oxidation reduction involves
equilibrium, it involves

723
00:43:42,030 --> 00:43:43,070
thermodynamics.

724
00:43:43,070 --> 00:43:46,020
I like it because it's really
important for reactions

725
00:43:46,020 --> 00:43:50,540
occurring in the body, and acid
bases as well -- p k a's

726
00:43:50,540 --> 00:43:52,480
are really important to that.

727
00:43:52,480 --> 00:43:55,860
And so, between acid base and
oxidation reduction, you cover

728
00:43:55,860 --> 00:43:59,220
the way a lot of enzymes work.

729
00:43:59,220 --> 00:44:01,720
So let me give you five minutes
of rules, and that

730
00:44:01,720 --> 00:44:03,950
will serve you well
in this unit.

731
00:44:03,950 --> 00:44:06,110
Some of these are
pretty simple.

732
00:44:06,110 --> 00:44:11,600
For free elements, each atom has
an oxidation number of 0,

733
00:44:11,600 --> 00:44:15,100
so this would be 0.

734
00:44:15,100 --> 00:44:19,720
So, oxidation number of
0 in a free element.

735
00:44:19,720 --> 00:44:25,830
For ions that are composed of
one atom, the oxidation number

736
00:44:25,830 --> 00:44:30,910
is equal to the charge of the
atom, so lithium plus 1 ions

737
00:44:30,910 --> 00:44:34,180
would have an oxidation
number of plus 1.

738
00:44:34,180 --> 00:44:37,730
Again, pretty straightforward.

739
00:44:37,730 --> 00:44:40,100
Group one and group two
make your lives easy.

740
00:44:40,100 --> 00:44:42,500
They seem to have a lot
of consistent rules.

741
00:44:42,500 --> 00:44:45,320
Group one metals in the periodic
table have oxidation

742
00:44:45,320 --> 00:44:47,070
numbers of 1.

743
00:44:47,070 --> 00:44:50,770
Group two metals have oxidation
numbers of plus 2.

744
00:44:50,770 --> 00:44:55,850
Aluminum is plus 3 in
all its compounds.

745
00:44:55,850 --> 00:44:57,970
Pretty simple.

746
00:44:57,970 --> 00:45:00,470
Now we get to things that are a
little more complicated but

747
00:45:00,470 --> 00:45:02,930
still useful, oxygen.

748
00:45:02,930 --> 00:45:09,020
Oxygen is mostly minus 2, but
there are exceptions to that,

749
00:45:09,020 --> 00:45:13,910
such as in peroxides where it
can have an oxidation number

750
00:45:13,910 --> 00:45:21,520
of minus 1, and if it's with
a group one metal, it

751
00:45:21,520 --> 00:45:23,200
can be minus 1.

752
00:45:23,200 --> 00:45:27,770
Remember, group one, and
actually group two here,

753
00:45:27,770 --> 00:45:31,770
that's plus 1, always plus 1,
always plus 1, always plus 2,

754
00:45:31,770 --> 00:45:35,050
and so hydrogen has to
accommodate that.

755
00:45:35,050 --> 00:45:39,260
So usually plus 1, except when
it's in a binary complex with

756
00:45:39,260 --> 00:45:44,090
these particular metals that are
in group one or group two.

757
00:45:44,090 --> 00:45:49,410
Fluorine, almost always minus 1,
or always minus 1 -- other

758
00:45:49,410 --> 00:45:55,100
halogens, a chloride, bromide,
iodide, also usually

759
00:45:55,100 --> 00:46:00,070
negatives, but if they're with
oxygen, then it changes.

760
00:46:00,070 --> 00:46:04,410
So, here is an example.

761
00:46:04,410 --> 00:46:07,130
And in neutral molecules,
the sum of the oxidation

762
00:46:07,130 --> 00:46:09,660
numbers must be 0.

763
00:46:09,660 --> 00:46:12,200
When the molecule has a
charge, the sum of the

764
00:46:12,200 --> 00:46:18,920
oxidation numbers must be
equal to that charge.

765
00:46:18,920 --> 00:46:21,630
So, let's do a quick example.

766
00:46:21,630 --> 00:46:28,620
Hydrogen, in this case,
is going to be what?

767
00:46:28,620 --> 00:46:34,340
Plus 1, so it's not
with a group one,

768
00:46:34,340 --> 00:46:35,690
group two metal here.

769
00:46:35,690 --> 00:46:40,830
So what does that leave
for nitrogen?

770
00:46:40,830 --> 00:46:46,490
And that makes the sum, plus 1,
which is equal to the sum

771
00:46:46,490 --> 00:46:49,530
of that molecule,
so that works.

772
00:46:49,530 --> 00:46:52,030
So we might not have known
nitrogen, but we can figure it

773
00:46:52,030 --> 00:46:55,130
out if we know the rules for
hydrogen and we know what it

774
00:46:55,130 --> 00:46:57,070
all has to equal up to.

775
00:46:57,070 --> 00:46:59,740
And so, this unit is sometimes
a relief after oxidation

776
00:46:59,740 --> 00:47:02,510
reduction, because it's all
about simple adding and

777
00:47:02,510 --> 00:47:05,390
subtracting, it's not so bad.

778
00:47:05,390 --> 00:47:08,780
OK, oxidation numbers do not
have to be integers.

779
00:47:08,780 --> 00:47:12,570
Example here, you have
superoxide, what would its

780
00:47:12,570 --> 00:47:16,170
oxidation number be?

781
00:47:16,170 --> 00:47:19,090
Minus 1/2.

782
00:47:19,090 --> 00:47:24,330
And those are the rules, and
then on Friday, we'll come

783
00:47:24,330 --> 00:47:27,290
back and we'll look
at some examples.