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CATHERINE DRENNAN: There
are five types of problems.
We just did two.
And now I'm going to convince
you that salt and water
problems are in fact the same
as weak acid weak base problems.
So we're going to move
now to today's handout.
So the pH of a salt
solution is determined
by what made that
salt. So a salt
is formed when you neutralize
an acid with a base or a base
with an acid.
And so depending on
what acid and which
base were used and
mixed together,
the salt might have
a different pH.
So an example here if
we had HCl and NaOH,
that's going to give our friend
table salt NaCl and water.
The pH of the salt and
water is not always neutral.
In this case it would be,
but it's not always neutral.
It depends on the
nature of the acid
and the nature of the base
that were mixed together
to form that salt. So
there are some rules
to help you figure out
whether something's
going to be acidic or basic.
And if you have a
salt that contains
a conjugate acid of
a weak base, so it
has a conjugate
acid in it, it will
produce an acidic solution.
And if you have a salt that
has small highly charged metal
cations like iron plus 3,
that will also be acidic.
So we saw last time I had
the prescription medicine
for my daughter, which was iron
sulfate, was highly acidic.
So note that periodic
table group 1 and group
2 ions, lithium plus 1,
calcium plus 2, sodium plus 1--
these are all
going to be neutral
and in fact, anything
plus 1 neutral.
And then salts that can take
conjugate bases of weak acids
will produce basic solution.
So you're going to
be asking yourself,
does this solution have a
conjugate acid of a weak base?
Does it have a conjugate
base of weak acid?
Does it have group 1,
group 2-- what's in there?
And then you can think
about whether it's going
to be acidic, neutral or basic.
So let's look at some examples.
So we have a salt, NH4Cl.
And we want to
figure out is this
going to produce an acidic
solution, a neutral solution
or a basic solution.
So we need to think about what
went in to making this salt.
And we're going to break
apart the salt and we're going
to think about NH 4 plus and
we're going to think about Cl-.
So let's think about
NH 4 plus first.
And we're going to ask
the question is NH 4
plus a conjugate
acid of a weak base
and therefore, a
weak acid itself?
One way to answer
this question is
to look up the Ka for
NH 4 plus, which happens
to be 5.6 x 10 to the -10.
You could also ask about the
weak base that it came from.
The weak base in this
case, its conjugate base,
is NH3 and ask if
that's a weak base.
So if you have something
that is an acid here,
you can think about
what the base would
be by removing H plus.
If you remove H plus from
NH 4 plus you get NH3.
And you can say, is NH3 a
weak base and look up a Kb.
In this case, it's 1.8
x 10 to the minus 5.
So what do you think?
Is this a weak base and this a
weak acid with those numbers?
What do you think?
Is that a weak acid?
Is that a strong acid?
So it's a weak acid.
Is this a weak base?
Yeah.
So we can look also in
this chart down here
and so NH3 is a moderately
weak base and NH 4 plus
is a very weak base, but
it is very weak acid,
but it is a weak acid.
So we do have a conjugate
acid of a weak base.
So it is acidic, not super
acidic, but it is acidic.
NH3 is a weak base
and so we probably
had NH3 as our base
that was being added
and it formed the
conjugate acid in this.
So then this would produce
an acidic solution,
because we have this
conjugate acid that's formed.
So from this part, just
from the NH 4 plus,
it should be acidic and
now we can consider Cl-.
So here we're asking
the question is Cl-,
a conjugate base of a
weak acid and therefore,
itself a weak base.
And if you tried to
look up the Kb of Cl-,
you would not find
it in any table.
But you could find in a
table its conjugate acid.
So if you add H to
Cl-, you get HCL,
and so you could look up a Ka
for that in the table and its
times 10 to the 7.
So is this a weak acid?
AUDIENCE: No.
CATHERINE DRENNAN:
It's a strong acid.
So it's not a weak acid,
it's a strong acid.
So is Cl- going to be basic?
No.
So it is not, it is not
going to be a basic solution.
If we look over here, we
see HCl is a strong acid
and so its conjugate base
is ineffective in it's base.
It just doesn't work
as a base at all.
So with a number of
10 to the seventh,
it really goes to completion.
It's a strong acid
by our definition.
And so Cl- is not going to
do anything to the solution.
It's not going to be
useful as a weak base.
So this solution is
going to be neutral,
or at least the part of
the solution due to Cl-
will be neutral.
So here we have something that
is acidic and something that's
neutral.
And so overall, the
solution is acidic.
So you need to break
down the two parts.
You had an acid and
a base being mixed.
Here we had HCl mixed with
NH3 and formed this NH4 salt.
And that's going to be acidic,
because HCL was a weak acid
and its conjugate
is ineffective.
Whereas, NH3 was a weak base.
So it has a
conjugate acid that's
a weak acid, a very weak
acid, but still a weak acid.
All right.
So that's how you think
about these salt problems.
So they really break down
to weak acid and weak base
problems.
So why don't you
give this one a try?
[AUDIENCE MUMBLING]
CATHERINE DRENNAN: All right,
let's do 10 more seconds
[AUDIENCE MUMBLING]
CATHERINE DRENNAN:
That is correct.
All right, let's look
at why that's correct.
So it will produce
a basic solution.
So we're going to
break it down to NA .
That's one of the
things in there.
And we ask is NA a conjugate
acid of a weak base?
Is it going to be
neutral, acidic or basic?
And what's it going to be?
It's going to be neutral.
It is not a conjugate
acid of a weak base.
It's group 1.
It'll be neutral.
So this is not
contributing to the pH.
So let's look at the second
part and look at CH3COO-.
So we have-- this can be
divided into Na+ and HC3OO-.
And then we ask is this a
conjugate base of a weak acid?
Or we could ask
the other question
is this over here a weak acid?
And the Ka for that was given.
So is that a weak
acid or a strong acid?
Is it weak or strong?
Weak, yes.
So the answer is yes.
It's weak and so it's conjugate
base would also be a weak base.
And so therefore,
this will be acidic.
So sometimes you'll be given Ka,
sometimes you'll be given Kb's.
If you're given
either one of them
you can answer the question.
If you want, you can
convert your Ka to your Kb.
What would you use
to convert Ka to Kb?
Kw, right.
But just with one of these
pieces of information,
you should be able to
answer this question.
So let's just look at
the general rule now.
You have a compound XY.
That can be broken
up into X and you'll
ask yourself is X a conjugate
acid of a weak base?
If yes, then the
solution will be acidic.
If no, it will be neutral.
For Y, you have Y-.
Is Y- a conjugate
base of a weak acid?
If yes, then it's going to be
add some basic nature to this.
If no, neutral.
And then over all,
you say, if it's
acidic plus neutral,
that's acidic, basic plus
neutral, basic, neutral
plus neutral, neutral.
And if I give you a
question of something
that is acidic and basic,
you can answer the question
without doing
math, because it'll
depend on the Ka of one thing
and the Kb of the other thing.
And you'll have to see
which one's a stronger acid
and which one's a stronger base.
And so you can't just
simply answer the question.
Now you may note,
that I did mention
that problem set 7 is long.
It takes a long time
to do these problems.
And so on the exam, writing
a fair exam when the problems
take a long time to do is hard.
But these are short
little things.
I could ask you to predict the
pH of salt by just thinking
about things and looking
up some Kb's and some Ka's.
So these are good short
questions on the exam.
So I'm probably not
going to give you one
if it's supposed to be
a short question that
is an acid and a base mixed
together, because then you'd
have to do some math.
So keep that in mind.
All right, so salts,
salt and water problems,
are really just weak acid
in water or weak base
in water problems, which
you already know how to do.
So we're going to move
on now to buffers.
Buffers.
So a buffer is a solution
that maintains approximately
a constant pH, there'll be a
little range of pH can change,
but the buffer tries to
keep the pH constant.
You use a buffer to
keep pH constant.
Why is this important?
Well, buffers are very
important in biology.
We all work at a constant pH.
Living things need to
be at a constant pH,
but a lot of times
chemical processes
also need to be
at a constant pH.
And here's just one example
of why buffers are important.
We're looking for
alternative energy.
We're trying to develop
alternative energy.
There's this idea of
creating fuel cells that
employ microbes.
These little red circles
are microbes here.
There are microbes
like shewanella
that will live,
adhere, to electrodes.
Will live on surfaces of
electrodes, metal electrodes.
And they will eat things
like sugar and organics.
And they will
respirate electrons
into the metal electrodes.
So the electrons go in and
you create an electric current
and you can do--
have a fuel cell.
So it's a way of generating
an electric current.
And then people have other ideas
that some like some microbes
live in electrodes, other
microbes like moorella
thermoaceticum, they
live on carbon dioxide.
They convert carbon dioxide to
things like acetyl CoA, which
is a fuel.
So you throw those in there
and then you like pump in CO2
and you make electric current.
So there's a lot of great
ideas of using microbes.
So some of my research
relates to understanding
the fundamentals
of these pathways.
And every once in
a while I venture
to hear one of these
big energy talks
about how is the development
coming of these ideas.
And so I heard one talk and I
was listening I wanted to know
is this really working?
Are people getting this to work?
What's the status?
And they talked about a lot
of technical difficulties
of getting it to
work, but they had
discovered something amazing.
Something that might really
change the ability to do this.
And it turned out
that when you're
generating all this
negative charge
you need to have positive
charge around, like H plus,
for example.
And so you need to
have protons, but it
mattered, the concentration
of those protons.
And they kind of had
to be kept constant
and they discovered buffers.
Fifty minutes of my life.
This was a talk at MIT.
The person discovered buffers.
I just felt like writing a
letter to Washington saying,
can I have their research
dollars because I already
know about buffers.
I don't need to
do a big research
project to discover that buffers
keep things at constant pH.
And that when you're
using living organisms
and doing experiments and
doing reactions, pH matters.
I already knew that.
Everyone who takes 5.111
already knows that.
And so this is part of the
reason I'm teaching you this.
Because some of you will not
go on and study chemistry.
You'll be doing engineering.
You'll be doing
alternative energy.
Doing all sorts
of things and you
can be the person in the room
that raises your hand to go,
it's called a buffer.
And that is going to change
the speed of that research.
They're just like
right away they're
going to be using buffers.
And when you do
Problem Set 7, you're
going to know how
to make a buffer.
So you cannot only
raise your hand and say,
it's called the buffer, you
can tell people how to make
a buffer to use in
their experiments.
So this is why the material in
teaching was really important.
I go out there and
I hear these talks
of people who are trying to
do these innovative things,
but they're lacking the
fundamental chemistry
knowledge that's hindering
their ability to do it.
So I want all of you
to have this knowledge
whether you study
chemistry anymore
or not, so that you
can be the person who
brings the case of
chemical principles
into the design process.
And I care a lot about
alternative energy.
It's going to be super
important in the future.
I care that we don't
destroy our planet going
after some forms of energy.
And this idea of using
microbes is a great one,
but we just need
people who know how
to work with them to
get it to go right.
All right, so buffers.
What are buffers?
You've got to know
about buffers.
A buffer consists,
an acid buffer,
consists of a weak acid
in its conjugate base.
They're often
supplied as a salt,
which means you have a counter
ion that comes with it,
like sodium plus or Cl-.
And it buffers on the
acidic side of neutral.
A base buffer or a basic
buffer has a weak base
in its conjugate acid,
often supplied as a salt,
and it buffers on the
basic side of neutral.
So the key to a buffer is
that it has a conjugate acid.
It has NHA.
But it also has
the conjugate base.
It has A-.
If it just has HA,
it's a weak acid.
That's not a buffer,
that's a weak acid.
If you just have A-,
you just have your base.
That's not a buffer.
You need to be both an acid
and a base, a base and an acid,
an acid and a base.
On the exam people
do buffer problems
and they just have one,
they don't have the other.
That's not a buffer.
And I'm going to write
on your little exam,
remember me twirling
around the room.
It was an acid and a base.
I don't think you
can ever forget this.
This will be in
your brain forever.
Buffer has an acid
and a conjugate base,
a conjugate base and
a conjugate acid.
OK, let's look at some examples.
So if you mix acetic
acid and acetate salt,
it's conjugate base supplied
in the form of a salt.
So here we have the acetic acid.
Here we have the acetate,
the conjugate base
often will be supplied.
There are probably going
to be some NA around.
And you get this
dynamic equilibrium.
It goes forward, it goes back,
it goes forward, it goes back.
So when you have
this, if you added
a strong acid to this solution,
you put in more acid into this.
And you had equal
amounts of this and this,
then you're conjugate
base is going
to react with the strong acid.
And you're going to go
in the back direction.
And you're going to neutralize
that acid that was added
and the pH should stay
more or less the same
if you did a good job
making your buffer.
So the buffer can respond
to the changes in pH
by shifts in
different directions
to use up the extra added
acid or the other way,
the extra added base.
So in this case, the acid that's
added is effectively removed
and the pH stays the same.
So a buffer can
respond to add acid.
A buffer can also
respond to added base.
If you add a base, a
strong base or any kind
of base, and in this
case hydroxide ion,
would then remove a
proton from the weak acid
from the acetic acid
forming water and some more
of the conjugate base.
And it's effectively
removed as well
and those OH ions are
effectively removed.
And again, if you did a good
job designing this buffer,
the pH should stay constant.
So that's why you need
the conjugate acid
and the conjugate base.
You need the acid to react
with the base that's added
and keep the pH constant.
You need the base to react
with the acid that's added
and keep the pH constant.
If you just have one, it's
not going to be a buffer.
So buffer action.
Again, weak acid transfers,
its protons, it's H plus, to OH
supplied by the base.
So here we have
added base and it's
going to be reacted with
the acid to be neutralized.
The conjugate base, A-, can
accept protons from a supplied
acid.
If you add an acid,
it will take those
and will neutralize
the acid that's added.
So let's think about what type
of acids and conjugate bases
are going to work to
make a good buffer.
A strong acid in
its conjugate base
does not make a good buffer.
Why don't you tell me why.
Yeah, so 67%.
So it's really about the
fact that the conjugate base
of a strong acid is
really not a base at all.
It's ineffective as a base.
And that's why it's
not going to work.
Buffer solution has to be
in a dynamic equilibrium.
You have to be able to push
it both ways, otherwise,
it's not going to work.
So a conjugate base of a strong
acid is ineffective as a base.
And so if you added acid,
it won't be neutralized
and the pH will change.
So you need to have something
that works as a conjugate base.
Something that's
able to interact
with the hydro ion or the
acid that's added and take
its proton to neutralize it.
If this doesn't happen,
the pH will be affected.
So you need to have weak acids
and weak bases in your buffers.
They could be moderately weak
or very weak or whatever,
but you have to have
a conjugate set that's
going to be able to push both
directions where the acid can
interact with the added base
and the base can interact
with the added acid.
All right, so all
of this is really
the same for a basic buffer.
So we can look at an
example here of ammonia
plus water going to ammonium
ion and hydroxide ion.
And we can think
about the same things.
When you add a strong acid, the
base will accept the protons
and make more of
your conjugate acid.
When a strong base is
added, the conjugate acid
will donate its proton.
And it will form NH3 again,
that conjugate base, and water
and the pH will stay the same.
So it's the same idea,
the only real difference
between an acidic buffer
and a basic buffer
is whether it pH's in the acidic
range or in the basic range.
But the buffers
work the same way.
So I can redraw this picture
to have our different symbols
on it, but the idea is the same.
The weak base will take
the proton supplied
by the acid and
the conjugate acid
BH is going to probate the OHH-
and again keep the pH constant.
So the idea again
of the buffer then
is that a buffer is a mixture of
weak conjugate acids and bases.
Again, they're weak
because its partner
needs to be effective
as an asset or a base.
It can't be ineffective.
Weak acid base mixtures
that stabilize pH
by providing a source
or a sink for protons.
It either adds protons
or takes protons away.
It can respond to add acid.
It can respond to add base.
So again, pH is important
and buffering is important.
Our body has its own
buffering system.
So we have carbonic acid
and bicarbonate buffering
agents in our blood.
And our blood has to be kept
in a sort of neutral range
over here.
If it gets too acidic, that
is very unhealthy for us,
leading to some
pretty severe symptoms
and too much leads to death.
If it's too basic,
that's really bad.
Also, we have death and
notice these are not
all that far away.
There are a number of
different medical conditions
that can affect
the pH, diabetes,
there are metabolic diseases.
Something else that is under
your control, hydration,
you need to drink enough water.
And as the weather
gets colder in Boston,
people so often stop
drinking as much water
or start drinking
like hot beverages,
which don't hydrate as well.
So keep hydrated.
If you're not hydrated,
if it's gets really bad,
it can start affecting the pH of
your blood, which is really not
good.
So buffering, very important.
All right, so let's do
a sample buffer problem.
So sample buffer problem--
and this is important.
I actually know of at
least one professor
that if you want to
do a UROP with them
or come to their
office and they're
going to say, write down how
you would design a buffer for me
and do those calculations.
So I can tell you later who
that is, maybe I won't, I
don't know.
But this is one of
the tasks that someone
uses to see if they
want you as a UROP.
OK, so here we have
one mol of formic acid
and 0.5 mols of the
conjugate base supplied
in the form of a salt
with sodium ions.
And those are added
to water and diluted
to a total final
concentration of one liter.
And you're given the Ka and
told to calculate the pH.
So the first step is to
write out the equation.
So we have an acid
and water going
to hydronium ions and a
conjugate base over here.
And we want to think
about what's there now
and what's at equilibrium.
So initial molarity, change in
molarity and your equilibrium
molarity.
So it's important
this word molarity.
Don't put mols in here,
put molarity in here.
But I made it
really easy for you,
because we have one mol and
one liter so the molarity is 1.
0.5 mols and one liter,
the molarity is 0.5.
And this is also
really important.
You're so used to only
putting things here.
When it's a buffer
problem, you've
got to put something here, too.
So you're adding the acid
with the conjugate base
at the same time.
This is not zero.
This is 0.5.
So now as this dynamic
equilibrium goes
you're losing some of this
and gaining some of these.
So at equilibrium we
have one molar minus x.
X is our hydronium
ion concentration,
which is what we want
to know to calculate pH.
And our conjugate base our
formic ion is 0.5 plus x.
We're given a Ka and we've
written this expression as acid
in water so we can use Ka.
The equilibrium constant is
for the expression as written
so we can put that in and
we can fill the rest out.
We have our products.
We have the concentration
of hydronium ion
times the concentration of
formic ion, the conjugate base,
over the concentration
of our formic acid,
and again water does not
appear in the expression
because it's the solvent.
Hydronium ion
concentration is x.
The conjugate base
concentration is
0.5 plus x and the
conjugate acid concentration
is 1 minus x.
So we can now try that
assumption that x is small
or we can use the
quadratic equation,
but why don't you just
try that assumption.
I'm going to take
this expression--
if you haven't written it all
down yet, put it right up here.
And now with the
clicker why don't you
tell me, if we use
the approximation
that x is small compared
to 1 and 0.5, what
does this simplify to?
All right, 10 more seconds.
So 70 something percent.
Yep, so let's take
a look at that.
So here we're making
the assumption
that x is small compared
to 0.5, compared to 1.
So that means that we drop
the plus x from the 0.5 term
and drop the minus
x from the one term.
So we're saying
that these are going
to be small enough
that it's still going
to be pretty much 0.5 and 1.
And then we have to
test that in a minute.
So if we use this,
we can calculate x
and it comes out to 3.54
times 10 to the minus 4,
and then we can
check the assumption.
And before you told me
how to check assumptions.
And that was I take
x and I'll divide it
by 0.5, which is the smaller
of these two, and times by 100%
and we get 0.69%,
which is less than 5%
so the assumption is OK.
And we don't have to
check it against 1,
because if the assumption x is
small compared to the smaller
number is valid, it will also
be valid for the bigger number.
So you only check it
for the smaller number
and then we can solve.
So we have to think
about-- we solved for x.
We're happy with x.
Our assumption was OK.
But then we have
to remember what
x is and x is the
hydronium ion concentration
and you always want to make
sure to think about this,
because if it's hydroxide
ion that's different.
So we can calculate pH.
pH is minus log of 3.54
times 10 to the minus 4,
which is equal to pH of 3.45.
And here we have two
significant figures
after the decimal point,
because the volume
we had before only had
two significant figures.
Everything else had three,
but the volume was just 1.0.
So that is our answer there.
All right, so I'm going
to just briefly start
what happens when you--
actually we'll do-- this
is how you would design
the buffer for this.
Next time we'll see what
happens if we stress
the buffer we design
by adding acid to it
and see what the new result is.
All right, so today we're
going to finish lecture 22.
So take out those lecture
notes on acids and bases.
And when we're done
with that, we'll
just continue with
acids and bases.
And we're going to
continue with some acids
and bases on Wednesday.
We should finish on
Wednesday and move on
to oxidation reduction and that
will end exam three material.
So exam three is
sneaking up on us
and it'll have thermodynamics,
chemical equilibrium,
solubility, and acid base.
So you should have
already filled
in your handout with
all the information.
The parts that you were
filling in are in bold here,
but I just want to remind you
of what we were talking about.
So we had a sample
buffer problem.
We had one mol of the
conjugate acid and a half
a mol of its conjugate base
supplied in the form of a salt,
were put in water and
diluted into one liter.
So we calculated what the
hydronium ion concentration
would be and therefore,
the pH, using information
that was given to
us about the Ka,
the acid ionization constant.
So we got a pH of 3.45.
Someone asked me after class,
but isn't this conjugate base
also reacting with water and
forming some of this weak acid.
And the answer is yes.
That we are writing this
as a problem of acid
in water, because
we can use the Ka.
But we could've also
written it the other way.
And written it, and this
is not in your notes,
but I just want to let you know,
that the weak base and water
could be written that way
forming a conjugate acid
and hydroxide ions.
And so then it would be 0.5
minus x and 1 plus x and x.
Use Kb then, because it's a
base in water to solve for x
and calculate pH from POH and
if you try it, you get 3.45.
So you can do these
problems either way.
You can think about
them as a weak base
problem or a weak acid problem.
You just have to remember
that this is not zero
when you start.
The conjugate has some amount.
That's what makes
it a good buffer.
So this can work either
way, but you also
want to remember if you write
it as an acid in water, use Ka.
If you write it as a
base in water, use Kb.
Because those
equilibrium constants
are telling you about those
ratios of the conjugate acid
and the conjugate
base at equilibrium.
So within the Ka and
the Kb is information
that you need about this
dynamic chemical equilibrium.
So both of these ways work and
by the end of, sort of halfway
through today's lecture,
you'll see yet another way
to solve buffer problems.
And probably will be for most
of you, the preferred way
to do it.
So there's a number
of different ways
which makes grading
the exam lots of fun.
And we have option 1,
option 2, option 3--
to make sure that all
the options are possibly
accounted for.
All right, so the
purpose of a buffer
is to keep the pH
pretty much constant.
So you can add a
strong acid to it
and the acid will
be neutralized,
keeping the pH constant.
You can add a strong
base to it and the base
will be neutralized, keeping
the pH pretty constant.
So what you want in a buffer is
to have a weak acid conjugate
base pairing that
allows it to be a source
or sink for a strong
acid or a strong base
to keep the pH constant.
So now in this
problem we're asked
to calculate what
would happen to the pH
if 0.1 mols of the strong
acid had been included
in our 1 liter solution.
And because 0.1 mols
of our strong acid
would react with
equal number of mols,
all of that conjugate
base, that's
what the conjugate
base does in a buffer,
it reacts with the strong
acid that is supplied.
And it will form equal
mols then of the conjugate,
its conjugate acid.
So we have to do
some subtraction.
So for the conjugate
acid, we had 0.5 mols.
For the conjugate
base, we had 0.5 mols.
We used up one of
that, 0.1 of that,
reacting with the strong
acid and we have 0.4 left.
So we have a new concentration
for our conjugate base, 0.4
divided by 1, keep the
math simple, 0.4 molar.
So for our conjugate acid,
we had one mol to begin with,
but we formed 0.1 more when
that strong acid was added.
So now we have 1.1 mols
and a new molarity.
So now we have to go back and
we can use this expression again
and determine what x is
now, what the hydronium
concentration is now using Ka.
So why don't you tell me
how I would solve this.
What is the correct
expression for Ka?
You can fill in your
little table as you do it.
All right, let's just
do 10 more seconds.
All right, so let's
just fill in that table
if you haven't already.
So here now, we can write this
as an acid and water problem.
Again, again, you could
have done the base in water
and used Kb, but we'll just do
it the same way we did before.
So we have our acid 1.1 molar.
Now remember, this is molarity.
So you don't want
to put mols in here,
you want to put a
concentration and some of that
will be minus x.
Then we have our
0.4, 0.4 plus x.
We can fill out the Ka.
The Ka again, weak
acid and water problem.
We have products-- hydronium ion
times our conjugate base over
our conjugate acid equals
0.400 plus x times x, these 2,
and 1.1 minus x on the bottom.
So you can always solve then
for x, which will tell you
about the pH by using the Ka.
And you just have to remember
molarity and remember
these are the new molarity
after the reaction has occurred.
All right, so we can
use the approximation
that x is small compared to
1.1 and 0.4 to simplify this.
Or we could use the
quadratic equation.
If we simplify it, we
calculate x as 4.87 times 10
to the minus 4.
And remember again,
that the approximation
is you get rid of the plus
x and the minus x here.
And then, you have to
check it with the 5% rule.
If you did that, if
you used the quadratic,
then you don't have to check.
But if you use the
5% rule, then you're
asking is this number 4.87
times 10 to the minus 4
within 5% of the smaller value,
which is 0.4 and it is here.
It's 0.12%.
So the assumption is OK.
That's less than 5%.
X is small.
And then we can
solve for the pH.
So pH is minus log of the
hydronium ion concentration,
which here is 3.31.
This only has two significant
figures after the decimal,
because if you remember,
back on the other page,
the volume only had two.
So actually all these
numbers here really just
have two significant figures.
We carried an extra, but
it really just had two.
All right, so our
buffer was pretty good
so we added 0.1 mols
of a very strong acid.
And we only changed the pH.
We only lowered it
from 3.45 to 3.31.
So that was a pretty good buffer
that we had designed there.
All right, so that is how
you do a buffer problem.
And there are other kinds
of things that you can
be asked about with buffers.
In particular,
you can also often
be asked to design a buffer.
And when you design
a buffer, you
must consider the
following things.
The ratio of your
conjugate acid base pair
and that they
should also be weak
conjugate acid base
pairs-- the pKa and the pH.
OK, so here comes a derivation.
So here is our expression
that should be familiar to you
at this point, for a
weak acid, HA and water,
going to hydronium ions
and our conjugate base, A-.
It should also be
familiar to you
that you can write
the equilibrium
expression for
this acid in water
as hydronium ions concentration
times the concentration,
the conjugate base,
over the concentration
of the conjugate acid.
Water, which is our
solvent, doesn't
appear in the expression.
We can rearrange this expression
now and solve for hydronium ion
concentrations.
We can take a log of
both sides, and get this,
log of hydronium ion
concentration, log of Ka,
plus log of HA over A-, multiply
by negative to get this.
And so we have
minus, minus, minus,
and minus the log of the
hydronium ion concentration
is what?
pH minus log of Ka is pKa
and this is the same as that.
So that gives us this expression
that pH equals pKa minus
the log of HA over A-.
Now it's important to point
out that these are equilibrium
concentrations of HA and
A-, because we derived this
expression from Ka and that's
an equilibrium constant.
So these are
equilibrium expressions.
And that makes this particular
expression slightly less
useful, because we
often don't want
to have to calculate the
equilibrium concentration.
We know how much we added
and so we often want
to use initial concentrations.
But we can consider whether
the initial concentrations
are actually pretty similar to
the equilibrium concentrations
i.e.
the x is small, less than 5%.
So we know that a
weak acid typically
only loses a fraction
of its protons,
hence the definition
of weak acid.
Weak acid in water
doesn't ionize that much.
And a weak base typically
only accepts a fraction
of the protons that it could.
It's a weak base.
So the initial
concentration is often
approximately equal to the
equilibrium concentration
and that's what
we've been finding
in a lot of the problems.
We've found that the 5%
rule works pretty well.
It's usually less than 5%.
X is usually less than 5% of
the initial concentration we had
of HA or A-.
And so therefore, we can
say that the pH is in fact,
approximately equal to the pKa
minus the log of the initial
concentration of your HA over
your initial concentration
of A-.
And this expression is known
as the Henderson-Hasselbalch
Equation.
And so this equation is only
valid when your hydronium ion
concentration is small
compared to HA and A-, i.e.,
less than 5%.
So this 5% rule must apply to
use the Henderson-Hasselbalch
Equation.
The Henderson-Hasselbalch
Equation
is a great equation to
use for buffer problems.
And so, we've
showed you a couple
different ways of
doing buffer problems.
This is the final way I'm
going to show you using
the Henderson-Hasselbalch
Equation,
but you can only use
it if x is small.
But most of the time,
x is going to be small,
because buffers are
only buffers when
you have a weak acid with
a conjugate weak base.
And when you're talking
about weak acids and bases,
x is small.
They don't ionize that much.
And importantly, and
I'll emphasize this,
this equation only
works for buffers.
Don't apply it for just regular
weak acid in water or weak base
in water, a strong acid in
water, a strong base in water--
it only applies for buffers.
Buffers, buffers.
Now I know that MIT
students love equations
and they love doing
math, and so they
try to apply this
equation to every type
of acid-base problem.
Don't do it.
Buffers, buffers.
You'll remember that now, right?
Because I'll make noises
again if you don't.
You remember that now, right?
All right, so let's use
the Henderson-Hasselbalch
to design a buffer then.
So say we want to design
a buffer of pH 4.6.
Say you're interviewing for
a UROP position and the UROP
supervisor, the PI
says, tell me how you
would design a buffer pH 4.6.
And you might go to the shelf
and see what was available
and then look up what
the Ka and the pKa are,
because a buffer solution is
most effective in the range
of the pKa plus or minus 1.
In fact, the closer the
pH you want is to the pKa,
the better the buffer
is going to be.
So we can look and see
what's in the range.
There are several that are close
to 4.6 that we can choose from.
And probably would
end up choosing
acetic acid, because that
would be on the shelf,
whereas, some of the others
of these would not be there.
All right, so acetate
is going to work.
It's a pretty common buffer.
So we can use acetic acid
with a suitable pKa of 4.75.
Then we can use the
Henderson-Hasselbalch Equation
because we're
designing a buffer.
So we can use that
equation and figure out
what the ratio of acetic acid
to its conjugate base should be.
So we can rearrange
this expression.
We know the pKa and we
know the pH that we want,
and so we can subtract those.
And we get that the log of
the ratio should be 0.15.
And then inverse
log will tell us
that the ratio then of the
acid to the conjugate base
should be 1.4.
So you should use
things with that ratio.
So for example, you could
use 1.4 molar of the acid
to one molar of
the conjugate base
and the total amounts are often
less important than the ratio.
The ratio is very important,
but if you go too low
in concentrations,
then that will
affect what's known as the
buffering capacity, which
is the ability of the
buffer to resist changes.
So if it's too dilute,
and a lot of strong acid
or a lot of strong
base is added,
then it won't be a
good buffer anymore.
So the ratio is very important.
The amounts are important such
that you have some resistance
to change.
So the higher concentrations,
the more resistant to change.
And also, if you use
too low a concentration,
the Henderson-Hasselbalch
Equation is no longer valid.
So you could be
asked to calculate
sort of what the
minimum concentration is
that you would need to use.
And so for a pH 4.60, you
can back calculate what
the hydronium ion
concentration is.
And it's 2.5 times
10 to the minus 5.
So for our 5% assumption
to hold or to be valid,
that this number, the
hydronium ion concentration,
is less than 5% of
either one of these.
The concentrations
here would have
to be greater than 5
times 10 to the minus 4.
So if we use one
molar or something,
that's way above this, but
that would be the minimum.
If it's less than that,
that 5% doesn't really hold
and you would not have a very
good buffer in that case.
It wouldn't be very
resistant to change.
X would be big compared
to those numbers.
So that's how you go
about designing a buffer.
And in that, there
were two things that
are really common mistakes.
When I'm reviewing a
paper for publication,
two of the things that
I see the most often
is that people one, do
not use the right buffer
for their experiment.
They say, oh, I'm at pH 8.
And they'll use a
buffer that is in fact,
not a good buffer at pH 8.
So who knows what their
data is telling us.
And the other thing
that people will do
is that they'll say, oh
yes, it was buffered.
And the pKa might be right, but
the concentration of the buffer
is so low that you
don't really imagine
the buffer's doing anything.
So now you know how to avoid
both of those pitfalls.
