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PROFESSOR: --
All the things would be relevant, so here are some
examples from a couple of lectures ago.
And so here we're thinking about what happens when we add
an inert gas.
And remember, it's all about the partial pressure.
So, you always have to ask yourself did the partial
pressure change, and partial pressure's going to change if
there's a change in volume.
So, the secret in this problem is realizing that if inert gas
is added, and the total pressure is kept constant,
what had to have happened?
Yup, the volume would have had to increase, and so the system
is put at stress, and it responds in a way to minimize
the stress.
So it's going to respond in a way to go from fewer numbers
of molecules to more molecules.
So, on one side of the equation reactants there are
three, and on products there are two, so it's going to
shift toward reactants.
All right.
So today we're going to have another clicker competition,
and because it is Halloween, this is the prize for the
recitation that has the most correct answers.
So, let's see how we do today.
All right.
So we're going to continue where we left off on
Wednesday, and so these are your notes --
I've added them to today's handout and they're also in
the handout from the last class.
And I want to make a note that it's a good idea to start this
problem-set early.
You don't know everything you need to know to do the
problem-set, but you do know a number of them, so there's
some questions on thermodynamics and
equilibrium, and Le Chatelier's principle, you can
do all those problems. So next week -- today we're going to
talk about bases and buffers, and then we're going to move
into acid-base titrations on Monday.
And so, the problem-set looks like it's not that long, it's
not that many questions, but the acid-base titrations have
many parts, and each part is actually quite long.
So it's a very deceptive problem-set.
So don't be fooled by the total number of questions.
All right, so we were talking about acid and water and base
and water and p h, and so we're going to continue with
base and water right now.
So here we have a base in water, and so in this case,
the water is acting as an acid.
It's giving up a hydrogen ion or proton to the n h 3,
causing it to form its conjugate, n h 4 plus, ammonia
ion, and also hydroxide ion.
So here we have a base in water, and when we're talking
about a base in water, we're going to talk about base
ionization constant, or k b.
So at the end of last class we talked about the acid
ionization constant, or k a, and when you're talking about
bases, you're going to talk about k b's.
So, k b, it's an equilibrium constant for this reaction of
a base in water, and so it'll be equal to the products in h
4 plus and hydroxide ion over the reactant, n h 3.
The water is the solvent here, and since this is all pretty
dilute, it's mostly pure and its concentration isn't going
to change, so it's not included in
that equilibrium constant.
So we have a k b when we're talking about a base in water.
And we were talking about a base in water, the equation
should work that you have hydroxide ions
on one side of it.
So here, the k b is 1 .
8 times 10 to the minus 5 at 25 degrees, so that's a fairly
small number.
And so the small value tells us that only a little bit of
the n h 3 is going to ionize when it's in solution.
So only a little bit it's going to form n h 4 plus and
hydroxide ions.
So that's what that small number tells us.
So that tells us that it's going to be a weak base.
So a strong base is something that's going to react pretty
much completely to go to hydroxide ion concentration.
A weak base only ionizes a little bit in water.
And you can tell about whether something's strong or weak by
its k b value, or if it's an acid, it's k a
value or p k a value.
So here are some general ways to write these equations.
We have a base in water and so the water's going to act as
the acid, base is going to accept that proton or hydrogen
ion forming a base, h plus, and hydroxide ion.
So the base is just written as b.
You could also write the base as a minus, something a minus
in water going to h a plus hydroxide ion.
So sometimes you might see that when you're talking about
a conjugate base of a weak acid.
So these are two expressions that you'll see that are
fairly generic that expresses what happens when you have
bases in water.
Now remember, you know it's a base in water, you better have
hydroxide ions on the other side, because a base in water
is going to be forming hydroxide ions, and acid in
water would be forming hydronium ions.
So, a strong base, again, almost completely ionizes to o
h when in water, and here we can know what's strong or weak
by the k b, so the larger the k b the stronger the base.
And like there is the p k a, there's also a term p k b.
P k b is minus log of the k b.
And the larger the p k b, the weaker the base.
Now you won't see p k b very much.
It's not used very much, most things are
converted to a p k a.
So you'll see p k a's quite a bit, and you will see p k a's
if you take organic chemistry, if you take biochemistry, if
you take biology, you'll be hearing a lot about p k a's as
we go along.
Not so much about p k a's, but p k b's.
And so what I want you to do when you hear about p k a's,
is remember that you've learned about it.
Because I have been confronted by some of my colleagues who
teach in advanced levels, and they said our students tell us
that you never talked about p k a's in freshman chemistry.
And I assure them that I did.
So I'll be emphasizing this.
And so I want to make sure that by the end of this unit
you're really familiar with p k a's because you'll need them
later on, and I want you to really impress my colleagues
in later classes, and they'll say oh, that's a 511-1
student, of course, they know what a p k a is.
Even my six month old daughter who's over there, she's like
what, people didn't know what a p k a is?
You don't want to get her upset.
OK.
So all of these things are related with the acids and the
bases, because for every acid it has a conjugate base, every
base is a conjugate acid.
And so, if you have a stronger acid, the stronger the acid,
the weaker its conjugate base.
And the stronger the base, the weaker its conjugate acid.
And this becomes very important in doing these
problems. So here's a little table that
emphasizes that fact.
So we talk about a strong acid.
Most people are familiar with h c l, hydrochloric acid.
So it's a very strong acid.
And its conjugate base, c l minus, is not really a base,
it's completely ineffectual as being a base.
It doesn't really do anything at all.
A strong acid really drives you all the way to hydronium
ion concentrations.
It doesn't go back the other way, it's not really
equilibrium, it's just going to completion there.
So the conjugate is really, really weak, basically not a
base at all.
Then we get into this middle range and here things that are
moderately weak or very weak acids also have their
conjugates in the weak range.
But if you get to something that is a very strong base
down here, its conjugate is going to be also ineffective
as an acid.
So if something is very strong, its conjugate is
pretty much non-existent in those properties, but when you
have weak-weak, then you can start talking about buffers,
which we're going to get into later in today's class.
OK.
So let's prove that, in fact, it has to be true that there's
a relationship between the conjugate acid and its
conjugate base or conjugate base and its conjugate acid,
that they both can't be strong.
One has to be -- you have to be weak-weak, strong or
ineffectual.
All right, so let's look at the first one up here.
So first let's look at what is this acting as?
What is n h 3 acting as, an acid or a
base in this equation?
So it's acting as a base.
And that means water is acting as an acid.
The water gives up a proton or hydrogen ion to the n h 3
forming the conjugate acid of that base.
And then the conjugate base is the hydroxide ion.
All right, so now let's write term for k, and so we're
talking about a base in water, so we're talking about k b.
So k b is going to equal what -- what do I put up here?
Tell me one thing to put up there.
Yup.
OK, n h 4 plus, and hydroxide ion over n h 3.
OK, so we don't have water in there.
All right, so let's look at the next reaction.
So what is n h 4 plus acting as?
It's acting as an acid here, so it's giving up its proton
or hydrogen ion to the water, which is going to act as a
base and accept that hydrogen atom.
And when this gives up its hydrogen ion or proton, it
forms its conjugate base, and the water is conjugate here,
is an acid, hydronium ion.
So we have our conjugate acid base pairs here.
So now am I talking about a k a or a k b?
I'm talking about a k a, so I'm talking about an acid in
water, and we know this is an acid in water if we look at
what's happening over here.
So we have an acid in water, and so we'll put our
concentration of hydronium ions and n h
3 over n h 4 plus.
So now we have k a's and k b's written for the conjugates,
the conjugate acid of n h 3 and n h 3 itself.
So now, we can think about what happens if we take these
k's and we multiply them out together.
All right, so we have a k a and we have a k b.
All right, so we have a k a and a k b, so if we take our k
a and times our k b, we're going to just multiply these
out together.
So I'll do this one first, k a -- just copy from above --
times k b, hydroxide ion, over n h 3.
And things are going to cancel out, and so I'm left with
hydronium ion and hydroxide ion.
What is this, when you have hydronium ion
times hydroxide ion?
What is that called?
It's another k.
K w.
So we just showed that k a times k b equals k w.
So we can take the logs of all our k terms here, and if we
take the log of k a plus the log of k b equal the log of k
w, or p k a plus p k b equals p k w equals 14.
So there's this relationship with a conjugate acid and its
base between its k a and its k b.
So if one is really big, the other has to be small or they
can both be sort of in the middle.
But they're always going to add up in terms of the p k a
and the p k b to 14.
And the thing about these problems is if you're given a
k a for an acid, you can calculate the k b for its
conjugate base, and you'll be doing that a lot in titration
problems that are coming up.
All right, so there's this relationship between the
strength of an acid and the strength of its conjugate
base, and let's just think for a minute again about this
concept of strong and weak, because this is really
important for the next unit.
So if we have a strong acid, h a, in water, it's going to go
pretty much completely over to hydronium ion and the
conjugate, and this conjugate is going to be really
ineffective as a conjugate base, as a base at all.
So it's going to really go all to that hydronium ion
concentration.
And so, in talking about a strong base, you don't really
have to worry about an equilibrium situation.
Just remember it goes pretty much to completion, and so you
can do complete subtractions when you're doing this.
And the same is true for a strong base.
So, for a strong base, any b, in water, it's going all the
way down, it's driving the reaction all the way over
here, and you're forming, you can consider it that however
much strong base you added is how much hydroxide
ion you have here.
How much strong acid you add is equal to how much hydronium
ion concentration you have here.
So however much of a strong acid or a strong base, you
think they go all the way to completion, but for a weak
acid we're going to have equilibrium, and so you'll
have to set up equilibrium tables to figure out if you
added this much weak base, how much did it ionize.
So remember that -- people get worried about the strong acid,
and you just assume it goes right to completion.
And you can tell again by the k a's and the k b's what's
going to be strong or not.
And so our definition for strong acids is that you have
a k a greater than one, strong base pretty much, the only
problems you use, people are adding sodium hydroxide or
potassium hydroxide, there are not a lot of options for
strong bases.
But for strong acids, people are always worrying about
whether they've identified those correctly or not.
OK.
So let's look at this relative strength of acid problem and
do an example here.
So in this equation we have acid in a base going on one
side, another different acid on the other side.
So we can look at whether the reaction is favored toward
this direction, toward the right or the left, depending
on which acid is stronger.
Will the reaction ride to the right or left.
So if this acid is stronger, then it should drive the
reaction this way.
If this is the acid on the other side, if it's stronger,
then you would expect to drive the reaction the other way.
So we can take a look at that.
So we can consider the k for the overall reaction, again,
just products over reactants here.
And we can also consider the reaction from just each acid
alone in water.
So we can consider it separately as well.
So first we can take a look at one acid.
So if we take a look at this acid alone in water, it'll
form h 3 o plus and n h 3 minus.
So when it gives up its hydrogen ion to the water and
then it forms its conjugate over here.
So now are we going to look at a k a or a k b?
K a?
So we're going to have our products over our reactants,
and the number is quite large, 20, for that k a.
And now we can look at the other reaction as well.
So we can look at this acid in water here and form your
hydronium ion concentrations and your conjugates -- again,
we're looking at that acid in water, so it's a k a, and we
have our products over reactants.
Now we have a number of 5 .
6 times 10 to the minus 10.
So, we can consider those two equations back together, and
this time we're going to be subtracting the equations from
each other to get our sum equation.
And because we are subtracting, we're going to
end up to dividing the equilibrium constants.
So when we add the equilibrium constants together, we
multiply things, and if we're subtracting we divide.
So the k in this case is going to be equal to the k a of the
first acid over the second acid, and you can prove this
to yourself, you write up the k a here, and the k a here,
and then some of your terms, the hydronium ions cancel, and
you get the k overall equilibrium constant that we
wrote in the beginning, again products over reactants.
We know the value for k a for one acid, we know the value of
k a for the other acid, and we can divide those to get the k
for the overall reaction.
And then you can tell me what that k means in terms of which
is the stronger acid of the two, and which side of the
equation does the reaction lie to, the right or the left?
All right, let's give 10 more seconds.
Yup, people did pretty well on this.
So you could have thought about it in terms of the
overall k or of the individual k a's.
So, the stronger acid is the one with the larger number.
And h n o 3 had a number of 20, so that was pretty big, so
that's a really strong acid.
And because it's a strong acid, it'll lie to the right,
so it's going to push toward products -- it's a strong
acid, so it wants to disassociate a lot, so would
push it that direction.
And you can see that also in terms of the equilibrium.
Overall equilibrium constant, if we go back to the slides
for a minute.
So this number overall, k, is also quite large, very large,
so that means a lot more products than reactants at
equilibrium.
So again, those are what you can determine if you're given
a table of k a values, which on the test you will get a
table of k a values, you can tell me a lot about different
reactions knowing that information about k a's.
All right.
So in this unit there are different types of acid base
problems, and sometimes it feels for people like there's
an infinite number of different types of acid base
problems. But, in fact, there are really only five.
And so one of the things I strongly recommend in this
unit and working problems is figuring out which type of
problem it is, and that will help you a lot in solving it.
So you can either have a weak acid in water, a weak base in
water, and sometimes you can be fooled and say oh, it's a
salt and water problem, but a salt and water problem
actually breaks down to a weak acid and water problem and a
weak base and water problem.
So it's really not a different kind of problem, and we'll see
that in a few minutes.
And you can have a strong acid in water and a strong base in
water, and then you can have my good friend the buffer type
of problem.
So those are the type of problems and being able to
recognize them is key to doing well in this unit.
So let's work a problem in the first type of a weak acid in
water problem.
All right, so what's a weak acid?
Well, vitamin C is a weak acid.
And so sometimes when you're taking your vitamins you get a
bad taste in your mouth.
And if you did take a vitamin C tablet here, which is 500
milligrams in this vitamin C tablet, and dissolved it in
water -- this is not scientifically measured -- but
dissolved it in water, so you would say taking your vitamin
with a lot of water, and it was starting to dissolve and
being pretty unpleasant, then we could calculate at
equilibrium what kind of p h we would have in that mixture.
Now, these vitamins, this Nature's Bounty, they do a
really good job of isolating the vitamin, so it is pretty
much impossible for it to dissolve.
So they have a nice coating around it that's highly
protective, at least at normal water p h's, so it doesn't
really dissolve.
And one year I thought I would do the actual experiment, we
could talk about significant figures, but I could not get
the tablet to dissolve --
I heated it, I stirred it, I did everything at neutral p h,
at room temperature.
I want to do it at room temperature, it just didn't
work, even the high temperature didn't work.
So if you buy Nature's Bounty, it will not dissolve in the
water that you're taking, so just for that little bit of
information.
But if you had, say, an inferior brand of vitamin C
that readily, that didn't have a nice coating around it, then
you could do this experiment.
So let's take a look at that.
So the first thing that we have to do is calculate the
molarity of the acid that we've added.
So here, just have to make sure that your units are going
to be correct.
So we have grams, we're converting it with a molecular
weight to moles, and then we have the number of moles in
the amount of water, and we can calculate the molarity of
that solution.
And one of the mistakes that people often make in doing
these problems, they forget to do all of the conversions that
are necessary, sometimes they stop at moles, and you're
talking about concentrations here, so don't forget about
your friend the volume.
All right, so then you can write and equation, and I
highly recommend that people do this on the test because it
helps them figure out what type of problem it is and it
voids people making silly mistakes.
So if we're talking about an acid in water, you should make
sure that your equation reflects an acid in water.
If you have hydroxide ion on the other side, something is
very wrong that's going on, and acid and water is going to
be giving hydronium ion concentrations
and a conjugate base.
So then we can set up an equilibrium table here.
We calculated the initial molarity.
And in the beginning there's nothing over on this side, so
we've just added our weak acid to water.
And so then the change, there's going to be some
amount of this that ionizes minus x, some amount of this
that's formed, and some amount out of the
conjugate that's formed.
So we have 0.0284 minus x plus x plus x.
Now we're talking about a weak acid in water, so what term am
I going to want to use next?
K a.
So I'm going to want to use k a next -- k a
value is here, 8 .
0 times 10 to minus 5, we have products over our reactant
here, and we have x squared over 0.0284 minus x.
Now you can make an assumption when you're working on these
problems and check it later.
So you can make the assumption that x is really kind of small
compared to this o.084, and you can just drop this x out
of the term here.
And then check later and see if that worked or not.
So that makes the math easier, and so now we can just solve
for x, and x comes out to be 0.00151.
The really two significant figures, but we're going to
carry an extra one for the moment, so they're just two
figures right here, two significant figures here.
Now we can check and see if x was really small, if that
assumption worked right.
So, is 0.0284 minus 0.00151 really the same as 0.0248, and
we let you make the assumption that it is, we say it's OK if
it's less than 5% of the value.
So, in this case, it's actually not, it's 5 .
3, so that violates our policy.
So it's more than 5%, so then you have to use the quadratic
equation to solve the problem.
I just want to note that this term, this percentage, can be
called sometimes percent ionized or percent
deprotonated, so that you're not thinking that's some kind
of bizarre term if you see that.
And if you use the quadratic equation, you get an answer of
0.00147, again, that's really two significant figures.
So it's not a whole lot different, actually, than the
number you got making the approximation.
So, once you know what x is, x is the hydronium ion
concentration in this problem, and so we can plug that in, p
h is minus log of this, so that's 2 .
83.
And so we had really two significant figures here, and
so we are going to have two significant figures after the
decimal point here.
And if you haven't reviewed your sig fig rules and need
more help if that seems wrong, then you should definitely
review it before the next test.
All right, so now we'll continue with today's lecture
notes, and we're just going to continue right on and we're
going to talk about weak bases, and well, we're start
working on our way through, we're also going to try to get
to buffers today.
So, we've done a problem for a weak acid in water.
So now let's talk about a weak base in water, and you can
start us off.
So in this problem we're given a molarity, so you didn't have
to calculate that, and now you can help me fill out the table
so we know what to do here.
OK, let's just do 10 more seconds.
Very good.
So you're going to be losing some of the amount -- some of
the amount of the weak base you have in, it's going to
ionize, and so then you'll be forming the conjugate acid
plus x, and you're going to be forming hydroxide ions plus x.
And so the one with the minus sign is important, and
sometimes there will be 2's involved, and that depends on
the stoichiometry of the reaction.
All right, so we can use that information now and go on and
look at -- actually you could leave that clicker question up
for a minute, and we're going to talk about the k b.
So the k b is going to be equal to our products, n h 4
plus and our hydroxide ion concentration over n h 3.
And so now I can fill in the values that you told me.
So we have, on the top we're going to have x squared, on
the bottom we're going to have 0.15 minus x.
And now we can make an approximation here that x is
going to be small compared to 0.15, and so we can say that's
just going to be equal to x squared over 0.15, and the k b
value that was given was 1 .
8 times 10 to the minus 5.
So now we can solve for x using this approximation, and
using this approximation x comes out to be 0 .
00164, and we can look at whether that
approximation was OK.
So is this number less than 5% of 0.15, so we can say 0.00164
over 0.15 times 100, and that comes out to be 1.1%, so
that's OK, that's less than 5%.
So that's good, we don't have to use the
quadratic equation here.
So now, we want to calculate the p h.
So can I just plug that number for x into my p h equation?
What is x?
What is x equal to here?
It's equal to two different things.
What's one of them?
Hydroxide ion concentration.
So what we can do is calculate a p o h.
So, p o h is minus log of the hydroxide ion concentration,
or minus log of 0 .
001647, and we really only have two significant figures
here, and that is going to come out to 2 .
79.
And so, we would have two significant figures after the
decimal point, because this number had two significant
figures in it.
But I'm not done.
I've calculated p o h and the problem wanted p h.
So how do I go from p o h to p h?
14 minus, yup.
So 14 .
00 at room temperature minus 2 .
79 is going to be equal to 11 .
21.
And so that makes sense.
Now in doing these problems always consider, sometimes
you're rushing and you get done you say OK, my p h is 2.
But go back and think about the type of
problem you're doing.
It's a base in water problem.
Would it make sense that the p h was 2 if it was a base in
water problem?
No.
And so then you realize oh, I have to do another step.
So that kind of thinking can save you a lot of points on
the exam, to remember what it is you're trying to calculate
and go back and make sure that your answer makes sense.
And sometimes people run into weird math problems and some
they'll write and say this p h should be above 7, it's 2, I
don't know what I did wrong, clearly I did something wrong.
I know that's wrong but I don't have time to figure out
what I did wrong.
That will get you points.
So just recognizing that if something makes sense or not
tells us you know what's going on, and sometimes math issues
can get you into a place that you can't get out of quickly.
So just thinking about whether the problem makes
sense is a big step.
All right.
So now, we're going to talk about salt problems and I'm
going to try to convince you that salts are actually the
same as the weak acid and weak bases that we just did.
So a salt is formed when you mix an
acid and a base together.
So, for example, if you have h c l and sodium hydroxide,
you're going to get table salt and a c l and water.
So the p h of a salt in water is not always neutral.
Sometimes it's neutral, sometimes it's not neutral.
Well, when would it not be neutral?
Well, if a salt contained a conjugate acid of a weak base,
then that conjugate acid is going to
make it weakly acidic.
Salts that contain things like iron 3
plus also may be acidic.
So when you're drinking water and you measure the p h of
that water and it's not neutral, this could be part of
the reason that there's some salt in the water, some
different ions in the water.
So a general rule from the periodic table group one and
group two metals, so lithium, calcium, sodium, those are all
going to be neutral in solution, so you can just
remember that.
And if a salt contains a conjugate base of a weak acid,
then it'll form a basic solution.
So it's all about whether the salt derived from a weak acid
or a weak base that's going to give you a clue as whether
it's an acid or a basic solution.
If it derived, say, from a strong acid mixed with a
strong base, then it's going to give you
a salt that's neutral.
So let's look at some examples.
So here we have n h 4 c l.
So we can break this down and think about where a salt like
this would have come from.
So it would have come from n h 4 plus, and it would have come
from c l minus.
So n h 4 plus, let's think about where this came from.
What is it?
So, we want to ask the question is n h 4 plus a
conjugate acid of a weak base?
And what is its conjugate base?
Well, it's conjugate base is n h 3.
If you lose a hydrogen ion or proton from n h 4
plus, you get n h 3.
And so you're really asking, if that's a weak base, then
its conjugate is also going to be weak.
So you need to know about these guys to see what would
happen if you have n h 4 plus in solution.
So how do you know about this.
Well, you know about things being weak or strong based on
their k a's and their k b's.
So, is ammonia a weak base?
It has a k b of 1 .
8 times 10 to the minus 5, so yeah, that's a small number,
so that's a weak base.
And so it's actually in this table, so if you have
something that's weak over here, then its conjugate is
also going to be weak over there.
These are totally lined up right.
And so the conjugate over here is also going to be weak.
And if you weren't really sure you could always look it up,
so here you have ammonium ion and it has a k a of 5 .
6 times 10 to the minus 10.
Yup, that's a very small number, that is a weak acid.
So yes, the conjugate is weak, the base is weak, and the
conjugate acid of that weak base is also weak.
So, n h 4 plus does have acidic properties.
It's not a strong acid, it's a weak acid, but it will make
things acidic.
So this should be acidic.
What about c l minus?
Do you think it's going to do anything useful for you?
Where do you think c l minus came from?
From h c l.
So we could ask again, is a c l minus a conjugate
base of a weak acid.
The acid is h c l, is that a weak acid?
No.
So we can look it up if you didn't remember it, 10 to the
7, definitely not weak -- very, very, very strong acid,
and if something is strong acid, its conjugate is
ineffective as a base.
So c l minus is ineffective as a base so it's going to be
neutral here.
So overall, you have something that's going to be acidic with
something that's going to be neutral.
So overall it'll be acidic.
So this particular salt and water is going to be acidic
because the things that were mixed together to get it, one
of them included a weak base, and so that's going to form a
weak conjugate acid, so it'll be acidic in solution.
All right, so let's look at another one.
I'm giving you the k a value in this problem, and knowing
that particular k a value, tell me what you think is
going to be true about this particular salt in solution,
whether it'll be acidic, neutral, or basic.
OK, 10 more seconds.
So, some people were a little fooled by the information I
gave you, so let's take a look at this.
So if we go to my presentation here.
So we can break this up into n a plus and c h 3 c o minus.
N a plus, is that a conjugate acid of a weak base?
Is that going to be acidic?
What do we know about things in group one, a column of the
periodic table.
They're going to be neutral.
Where do you think that this came from, n a plus, where
might it have come from?
It might have come from n a o h, that might have been a base
that was added.
So it's not going to do anything for you -- things in
group one and group two are going to be neutral.
All right, so we can ask the question about
a c h 3 c o o minus.
Is it a conjugate base of a weak acid?
So then we can say is the acid that it came from a weak acid?
Is its conjugate a weak acid?
Well, how do we know about that?
Well, we know about that from the k a value that I gave you.
So is this a weak acid?
Yes.
So is its conjugate going to be a weak base?
Yes.
So, given that information we can say yes.
So its conjugate acid is weak, so then it
would be a weak base.
So if it's a weak base then it'll be basic in solution,
and we have something that's neutral plus basic, so overall
you get basic.
All right.
So let's look at a general example now of this as well.
So here we can talk about a general rule.
So if you have compound x y, we can talk about
x plus and y minus.
And for x plus you're going to be asking about is it a
conjugate acid of a weak base.
For y minus you're going to be asking is it a conjugate base
of a weak acid.
So for the first part you're asking about if it's a
conjugate acid, in the second half you're asking about if
it's a conjugate base.
So if something is a conjugate acid of a weak base, and
that's yes -- if you know that's a weak base or you know
that it's a weak acid, then it's going to be acidic, if
it's not, it'll be neutral.
Same thing is true over here.
You might know about that something is a weak base, you
might know that its conjugate is a weak acid, and if you
have a conjugate base of something that's a weak acid,
then it is a base.
If it's a strong acid it would be ineffectual, but if it's a
conjugate of a weak acid, it'll be basic.
No, it's neutral.
So overall, you can have three possibilities.
Acidic plus neutral is acidic, basic plus neutral is basic,
and neutral plus neutral is neutral.
Now some people might come up with another option here.
What other thing am I leaving off of this overall?
Acidic plus basic.
That's because I'm never going to ask you that when
it comes to a salt.
Because pretty much, salts are formed when you're doing a
titration, and you're going to be either titrating a strong
acid, a strong base, you're going to be titrating a weak
acid with a strong base, or you're going to be titrating a
strong acid with a weak base.
You are never going to titrate a weak acid with a weak base.
That would yield no interesting
results of any kind.
So you're not going to be forming salts that are
conjugates of both those things.
So, if you want to think about it that way that's fine, or
you could just remember that that is not what I'm
going to ask you.
These are great little short answer questions on an exam,
so if you're good at thinking about this, it'll definitely
give you a couple of points on one of the exams.
All right, so now and last, just a couple minutes, I just
want to introduce very briefly buffers.
So a buffer is something that maintains the p h of a
solution, so it's going to buffer that solution.
So if you add a little bit of strong acid or a little bit of
strong base, it doesn't matter the p h is
going to stay the same.
So there are two kinds of buffers.
You have an acid buffer, which is going to buffer/maintain
the p h on the acidic side of neutral.
And a basic buffer, which will maintain the p h on the basic
end of the p h scale.
So let me just give you a brief example of a buffer and
just get you thinking about buffers.
So here, in about a buffer problem you're going to mix an
acid with its conjugate base.
So, you have acetate, and then probably the acetate salt of
the acetic acid.
So over here, you have the acetic acid, on this side you
have its conjugate base usually added in
the form of a salt.
And then you have an equilibrium.
So, what's going to happen if you add a strong
acid to this solution?
If you add strong acid, if you add more h 3 o plus, what
happens if you add more?
What direction will the reaction shift?
So you'll get the back reaction, it'll try to
minimize that stress and move the other way, and it'll use
up some of that acid and maintain the p h.
Then you can think, so these amount of acid added is
effectively removed and the p h stays the same.
What about if you add a strong base?
Well, that strong base will react with the acid, it will
remove protons from this acid, or the hydrogen ion here,
forming this water and its conjugate.
So you'll make more of these and the p h will
also stay the same.
So the base is going to be removed by reacting.
So they're effectively removed and the p h stays the same.
So in this you have a weak acid, h a, it'll transfer
protons to o h minus supplied by the strong base.
The conjugate of that weak acid, which would be a weak
base, is going to accept protons from any acid that is
going to be supplied.
So, in this way, you maintain the p h.
And so I just to emphasize that in a buffer solution you
have h a, acidic buffer solution, you have h a, and
you have its conjugate.
And if you only have one or the other, it's not going to
make a good buffer.
So, I want you to remember that in buffers you have both
conjugates -- one alone is not going to work.
And people forget this in the class, and so you can remember
that for Halloween, your chemistry professor dressed up
as a buffer to help you remember that in a buffer
you're going to have both.
You've got to have the conjugate acid base set,
otherwise it will not buffer.