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OK we are ready to start titrations
between an acid and base where one of them will be weak.
We will never do it when both are weak
but where one of them is weak and one of them is strong.
So we are going to begin here with a weak acid
and a strong base titration.
Acetic acid
is the first substance mentioned here.
We look at here and this is a weak acid.
It is not on our strong acid list, it is a weak acid
and sodium hydroxide is a strong base.
So we will begin by writing the reaction
between the weak acid the the strong base.
I will never include the sodium ion it is just
spectator ion, and we will write the net ion
equation for this reaction.
This is the biggest challenge for students is to write
a good reaction and know how to work the problem
from there. If you have got a good starting point, it is not a problem.
So we write out acid, we write OH- for out
base and we do a proton swap.
The acid donates the H+
to the HO- and produces the water.
We are left with the conjugate base of Acetic acid.
Now once this equilibrium occurs
then we have a point where
all of the acid has reacted with all of the base.
Then we have produced this conjugate base in solution.
This conjugate base which is the acetate ion
will undergo an equilibrium in water.
Now this reaction of it in water is at
equilibrium so we put the equilibrium arrows
again it is just a proton swap.
The water is behaving as the acid
and it will donate one of its protons
to the anion here to produce
acetic acid and hydroxide. to the anion here to produce
acetic acid and hydroxide.
So at the equivalence point we have this weak
base that has been produced.
What do we mean by equivalence point? Well we have just
consumed all of the acid and all of the base.
We have produced this conjugate base, the acetate ion
it is a weak base
it will re-establish an equilibrium with its
conjugate acid in water.
Because of this reaction here, this solution
is basic at equilibrium.
OK we are going to look at the titration curve
of a weak acid strong base titration.
Now I have changed my weak acid to of a weak acid strong base titration.
Now I have changed my weak acid to
formic acid, HCOOH but it is a weak
acid again. I am titration it with a strong base.
So we have ourselves a little flask and in this little flask
there will be an acid, and it will be acidic.
And we that at the very beginning here were we are
just looking at the pH of this weak acid.
That we have an acid pH.
If we wanted to calculate the pH of that
acid we would work a
problem very similar to what we did in the last unit.
How to calculate the pH of a weak acid.
So that is at the very beginning. Then we have our burette.
And in our burette with have a little with a little stopcock here we put the sodium hydroxide.
We start adding a little bit of sodium hydroxide.
We put a few drops in, a few drops
in here of our sodium hydroxide. Let consider We put a few drops in, a few drops
in here of our sodium hydroxide. Let consider
what happens, look back up here at the reaction. If you add a little
bit of the hydroxide to
a bunch, a lot, of the acid. bit of the hydroxide to
a bunch, a lot, of the acid.
Then the little bit of that base is going to get a bunch, a lot, of the acid.
Then the little bit of that base is going to get
consumed and go away.
And we will not have any of that base
and we will have converted some of this
HCOO- and produced some
of the HCOOH. So this will be consumed
and we won't have any
and we will have some of that and we won't have any
and we will have some of that
and what do we have? Well we have some of both. Well we have a weak acid
and its conjugate base in solution.
When you have a weak acid and its conjugate base and its conjugate base in solution.
When you have a weak acid and its conjugate base
in solution you are going to have a buffer.
So we when you are prior to this equivalence point
you are going to be in a buffer range.
You will not here that it kind of jumps up
and then it sort of levels off before it rises again.
We are resisting the change of pH
during that range there. Well eventually we will get to the point where
we have added
exactly the same amount of base
to consume all of the acid.
And we will have converted all of the acid over to
the formate ion. Well what
do we have there?
We have a weak base.
This is the weak base present there.
And so we should have a pH above 7 which it does.
So we have just consumed all of the acid and
all of the base that was added into the flask.
We have a conjugate base an a weak acid and we would
calculate the base at that point
based upon that weak base.
Now if we keep on adding
hydroxide beyond that point
we will produce, we won't have anything for it to be neutralized by.
So at that point we will have a strong base we will produce, we won't have anything for it to be neutralized by.
So at that point we will have a strong base
and the weak base in solution
and we so we would calculate
based off of this strong base
because it would really dominate how to calculate the pH.
So that is what a titration curve looks
like when you are doing a weak acid and a strong base.
So now we are ready to consider an acid being strong
and the base being weak. This is still apart of learning objective 6
where we are doing a titration where one of them is weak and one of them is strong.
We are going to consider this reaction here HCl and Ammonia
and do a titration between these two substances.
When i write this reaction
I will always write
the strong acid as H_3O+.
Not as HCl.
I want to get rid of the spectator ion,
which causes confusion when you are trying to write the reactions.
So anytime we do problems
with acids and base I will write H_O+
to be my strong acid. Lets just think why that would be?
HCl is in water
and when it is in water it will 100%
ionixe to give you H_3O+ and when it is in water it will 100%
ionixe to give you H_3O+
and Cl-. Well, Cl- is just a spectator ion ionixe to give you H_3O+
and Cl-. Well, Cl- is just a spectator ion
so whatever the concentration of HCl is in there that is
concentration of H_3O+ and that is what I will write
when I write my reaction. So I have a strong acid and a weak base.
And I do not have to memorize the products I just think about the
acid donating to the base. So the H+
that is on this acid, will be donated to the ammonia
H_3O+ becomes water
and ammonia becomes ammonium NH_4+.
Now when we are at the equivalence point and we have NH_4+ formed
and we have just consumed all of the acid and all of the base Now when we are at the equivalence point and we have NH_4+ formed
and we have just consumed all of the acid and all of the base
this NH_4+ will then need to undergo
equilibrium with water. this NH_4+ will then need to undergo
equilibrium with water.
So lets have you think through what NH_4+ should do.
First of all, what is NH_4+ stop and answer.
If you said it was an acid, you are correct.
It is positive charge and a H+ that it can donate.
So if it is an acid
when we to think about the reaction that
acid will do in water.
And you have to be able to write that reaction.
We write NH_4+.
We place it in water We write NH_4+.
We place it in water
we know we will have to use equilibrium arrows We place it in water
we know we will have to use equilibrium arrows
because this is a weak acid. we know we will have to use equilibrium arrows
because this is a weak acid.
We will do the proton swap, we won't memorize
products. We will say the acid will donate to the base.
And if the acid donates it is going to form it conjugate
base NH_3 and the water will
accept that proton and become H_3+
So this is an acid, it behaves as an acid
it will give us a pH
less then 7 at the equivalence point.
So lets look at this titration curve.
We beging with our flask
and in my flask is the ammonia, the weak base.
So it has a high pH. There is the pH. and in my flask is the ammonia, the weak base.
So it has a high pH. There is the pH.
Depending on what the concentration of that base is
it will have some pH.
We start adding the HCl.
So lets write the reaction that is going on again.
We have our NH3
our H_3O+
we are pushing this to completion and we have
NH_4+ and water.
So we are adding from out burette
the HCl.
It is going into the flask
and the first think that is going to happen is
some of your ammonia is going to be converted into
to ammonium.
We are going to have a ridge in here once again
that is a buffer range.
This is our buffer range.
Now something that I didn't mention on the previous
acid base titration, where one is weak and one is strong Now something that I didn't mention on the previous
acid base titration, where one is weak and one is strong
is what happens when you are halfway to your equivalence point. acid base titration, where one is weak and one is strong
is what happens when you are halfway to your equivalence point.
When you are halfway to your equivalence point the pH is
going to be equal to the pK_a
of the acid, the weak acid in solution.
Well, the weak acid in my solution of the acid, the weak acid in solution.
Well, the weak acid in my solution
is the ammonium ion.
So the pH is going to be the pK_a of ammonium.
Lets think about why halfway.
When you are halfway
to the equivalence point you will have converted
half of your ammonia to the equivalence point you will have converted
half of your ammonia
over to ammonium.
And you will have equal amount of both the ammonia
and the ammonium. And if we are a buffer we know that the pH
equals pK_a plus the log and the ammonium. And if we are a buffer we know that the pH
equals pK_a plus the log
the concentration of the base
over the concentration of the acid.
When you are halfway to the equivalence point
the amount of base and the amount of
acid are equal the amount of base and the amount of
acid are equal
therefore the log of 1 is 0
and that gives you pH equal to pK_a.
So that is where you will be always at halfway to the equivalence point. and that gives you pH equal to pK_a.
So that is where you will be always at halfway to the equivalence point.
When you are at the equivalence point however
you have formed only ammonium.
So the pH would be below 7. you have formed only ammonium.
So the pH would be below 7.
In this problem, the pH is 5.26
at the equivalence point.
It is acidic, because ammonium is acidic
and you would calculate the pH
by doign a problem based on
a weak acid in water.
So we are going to consider
a problem here where we are going to calculate the pH
when 10 ml of sodium hydroxide
is added to 30 ml of 15 molar
and that is formic acid.
You can't do these problems unless you can write a good reaction. Now I want you to come up with
the net ionic reaction between the acid and the base.
Stop and think about those substances
and choose which one would be the net ionic equation.
Well, hopefully you picked number three.
But if you did not lets talk about it, because it is the most important Well, hopefully you picked number three.
But if you did not lets talk about it, because it is the most important
step. A lot of students can do the calculations just fine
but they don't know how to start the problem. step. A lot of students can do the calculations just fine
but they don't know how to start the problem.
We consider the sodium hydroxide
and we have to think about what that is.
It is a strong base
I know it is a strong base because it has that OH-
a metal with OH- in there.
I get rid of the sodium because it is a spectator ion and I write OH-. a metal with OH- in there.
I get rid of the sodium because it is a spectator ion and I write OH-.
I think about HCOOH.
And I see that COO portion
I know it is a weak acid.
Organic weak acid and it is going to donate
that H at the end.
So this is a weak acid.
Now you don't memorize the products you just think
acid and base will swap the H+.
The H+ is this H.
Not the one on the left.
It is going to donate to the OH-.
If the OH- takes on an H+ it is going to make water.
And if the acid donates If the OH- takes on an H+ it is going to make water.
And if the acid donates
its H+ it is left with its conjugate base And if the acid donates
its H+ it is left with its conjugate base
ion right there. So we did an proton swap and formed those products.
So now we ready to do the calculation.
We start we our reaction OH-
plus HCOOH.
We have our products
and since it is a one-way reaction I am going to do an ICF table.
With an ICF table I am going to
figure out the moles
because that is what you put into an ICF table is moles. figure out the moles
because that is what you put into an ICF table is moles.
The moles of hydroxide.
Well since there is one hydroxide
in sodium hydroxide it would be the same as the moles
of sodium hydroxide.
This is going to be molarity
of the sodium hydroxide
times the volume
of the sodium hydroxide in liters
and that is going to give me
.0010 moles.
And that is what we would place here.
For the acid
again we will take the moles.
To get the moles of the acid, we will take the molarity of the acid again we will take the moles.
To get the moles of the acid, we will take the molarity of the acid
times the volume of the acid it is 30 ml or
.03 liters.
And that is going to produce .0045 moles
of the acid. So we will put it here.
We don't have any of the formate ion yet.
And I do not care about the water in my table.
An ICF table I consume until I consume And I do not care about the water in my table.
An ICF table I consume until I consume
the smallest quantity. So I am going to
consume .0010.
And produce those quantities.
So that will have consumed all of the strong base.
I will still have this quantity of weak acid
and I will have quantity of its conjugate base.
Well that is a buffer.
And as a buffer we will use
the Henderson-Hasselbalch equation.
pH equal pK_a
plus the log, and since the table is in moles,
we will just use the number of moles of base plus the log, and since the table is in moles,
we will just use the number of moles of base
over the number moles of acid rather then we will just use the number of moles of base
over the number moles of acid rather then
going ahead and converting them to concentration.
The pK_a is the negative log of the
K_a which is given there at the top of the screen.
And we have to consider which one is the base.
This is the base, we put it on top.
This is the acid, so we put it one bottom. This is the base, we put it on top.
This is the acid, so we put it one bottom.
And this will give me
3.77 minus And this will give me
3.77 minus
0.54
and the pH would be 3.23.
So we are before the equivalence point, we are in a buffer and the pH would be 3.23.
So we are before the equivalence point, we are in a buffer
that is buffer based upon an acid
and its conjugate base so it should
be acidic and that is the pH. and its conjugate base so it should
be acidic and that is the pH.
So now we are going to continue on.
We are going to go to the equivalence point. So now we are going to continue on.
We are going to go to the equivalence point.
To be honest, calculating the pH at the equivalence point
of an acid base titration is the most
difficult. It is not
impossible. I wouldn't be teaching it, if it was impossible
but you have a lot of work to do.
And you might not notice that you are at the equivalence point
sometimes until you are working through the problem. This problem
is stating right up front that we are at the equivalence point.
So lets see why this is challenging. is stating right up front that we are at the equivalence point.
So lets see why this is challenging.
Well we begin the problem as we did as before.
We have the same reaction, it is a strong base
and a weak acid.
One-way reaction
to produce the conjugate base and water.
We do out ICF table.
Now we don't know how much we have in there yet. We do out ICF table.
Now we don't know how much we have in there yet.
but we do know how much acid is in there.
And this is the same amount that we had before
.0045 .
So that is not changing.
now we are continuing to add base until we reach the equivalence point.
By definition the equivalence point is zero.
Is going to be the same moles
of the base as the acid we have put
in the same exact number there.
So our line looks like this so we have
added enough base to consume all of the acid.
And so we going to subtract both of these numbers.'
And so they are the same and produce on this side And so we going to subtract both of these numbers.'
And so they are the same and produce on this side
and that will have consumed
the reactants, and i only have
a weak base
We have a weak base in solution.
And we have only a weak base in solution. We have a weak base in solution.
And we have only a weak base in solution.
So anytime you are at the equivalence point this is what it is going to look like.
You are going to have zeros
between these two because at the equivalence point you will
just consumed all of your weak acid
with a strong base in this case.
And you will have only a weak
conjugate sitting over on this side. And you will have only a weak
conjugate sitting over on this side.
And so this is a weak
base, we need to write the weak base reaction.
So we place that weak base, knowing it is in water, we react it with water.
We do equilibrium arrows because it is weak.
And we proton swap we take the
H off the water and put it on the base.
The base is a proton acceptor.
The water is left behind as an OH-.
Since this is at equilibrium we don't do an ICF
table, but we do an ICE table.
And we have to put in, molarities into this table.
Where this was moles.
So we have to do a little work in order to figure out
the molarity. What we have to know is the total
volume of that solution.
Well we know we have 30 ml
of this solution. Well we know we have 30 ml
of this solution.
What we don't know is how many milliliters of the
base we had to use in order to get to the equivalence point.
So, keeping in mind we are trying to calculate base we had to use in order to get to the equivalence point.
So, keeping in mind we are trying to calculate
the molarity of
the base.
We know how many moles we have
we have to divide by the volume in liters.
We know we have 30 ml or
.30 liters of the
acid. So to figure out the liters of the base
we are going to have to
consider the fact that we
added this many moles of base
and we know the molarity of the base.
So if we know moles and molarity we can get the
volume of the base we added.
The moles of the base was .0045 that we had to
add to our flask to reach our equivalence point.
What I want to know is the volume
well the molarity of the base was .1 mole per liter.
That is going to give me 0.045 liters.
And I can put that here.
So I now I know the total I know the moles of
the base in there and I know the total volume that is in there.
And I can divide those numbers
and I will get a molarity
equal to
0.060 molar.
And that is will go in this table.
I don't care about the water.
Before the equivalence point this is 0.
And Hydroxide we will put 0.
Although we know that in water its
1 x -7
but it is small enough to consider it a 0.
I will consume all of this. Produce
x's on this side.
Give me 0.60 - x.
X and X [student don't have trouble filling in that table]
you have a lot of practice doing an ICE table.
But at this point we are ready now to
set it equal to our K expression.
Well we have to keep in mind that what we set it equal to our K expression.
Well we have to keep in mind that what we
got sitting right here is a base. This is a base reaction
so we are going to need K_b.
And K_b is going to be the HCOOH concentration
times the OH- concentration And K_b is going to be the HCOOH concentration
times the OH- concentration
over the HCOO- concentration.
Well, K_b, we weren't given, but we were give K_a. over the HCOO- concentration.
Well, K_b, we weren't given, but we were give K_a.
And K_b is 1.0 x 10 ^ -14 Well, K_b, we weren't given, but we were give K_a.
And K_b is 1.0 x 10 ^ -14
over the 1.7 x 10 ^-4.
And that is going to be equal to x squared [x^2] over
0.060 - x .
Now I recommend you stop
the video and work through and solve for x.
And then resume it once you have an X. the video and work through and solve for x.
And then resume it once you have an X.
Well did you get an x of 1.9 x 10 ^ -6 ?
That is what you should have obtained.
Once you have determined that you would think
I have determined X what does X represent?
Well here is X.
And X represents the OH- concentration.
So if I took the negative log of that value
that would give me the pOH.
The pOH
equals 5.73
and how do we get the pH? Well, the pH
equals 14
minus the 5.73.
And that is going to be a pH of 8.27.
Now lets stop and think if this makes sense.
Should the pH at the equivalence point
be a basic pH?
Well certainly, because we produced a base.
That is the substance in this solution
when we have reached
the equivalence point.
So having a pH above 7 is what we should see.
Here is a calculation in which
we have a strong acid
and a weak base.
Now I have not given you the formal for that weak base
it is actually HC_3NH_2
but if you do not know the formula for your base you can
either look it up or use a B to represent a weak base.
Now we are going to calculate the pH
to the equivalence point between those, so we are going to write a good reaction.
We have H_3O+ as our strong acid
and here is my weak base We have H_3O+ as our strong acid
and here is my weak base
and it is a one-way reaction because the strong
forces it to completion.
The acid donates to the
base and turns the base into it
conjugate BH+.
And the acid becomes water.
We will do an ICF table
but before we do this, and we are going to do this at the equivalence point.
Lets think about what the pH at
the equivalence point should be. Lets think about what the pH at
the equivalence point should be.
At the equivalence point we will just have consumed
all of the acid and all of the base.
And we would have converted it to BH+ and BH+ is what?
It is an acid.
And as an acid it should have a pH below 7.
So we will keep that in mind and make sure we have a pH below 7 when we are finished.
So we need to put into our ICF table what?
Well, moles. So we have to put mole in.
Now they did not give me
a volume to use, so we can choose any volume.
The pH at the equivalence point is the same whether you a bit a volume to use, so we can choose any volume.
The pH at the equivalence point is the same whether you a bit
vat of it, or a small quantity of it.
So I am just going to choose a quantity.
I am going to choose that I am going to have
in my flask, 10 ml
of that acid.
Now if we are at the equivalence point we are going to have equal moles
not equal volume, not equal molarities but equal moles. Now if we are at the equivalence point we are going to have equal moles
not equal volume, not equal molarities but equal moles.
So we will worry about the volume of the not equal volume, not equal molarities but equal moles.
So we will worry about the volume of the
base that we need to add
here in just a minute. But lets take the 10 ml
of the strong acid.
So to obtain the moles of H_3O+ of the strong acid.
So to obtain the moles of H_3O+
it would be the same as the moles of that strong acid.
And that would be the molarity
times the volume, I and I choose 10 ml.
And we will have
0.0010 moles
of the strong acid.
And we can place that into our table.
Now that we can know
that if we are at the equivalence point we have the same moles
of that substance.
For the equivalence point we have none of that
and the water doesn't matter.
We consume all of
of the BH+.
So we always have at the equivalence point
zeros on the left side.
and we have 0.010 here of this weak acid.
When we are at this point in our titration
you might want to go back and look at the procedure to
follow and consider what we have to do next. you might want to go back and look at the procedure to
follow and consider what we have to do next.
If it is a weak acid only
then we have to follow this
ICF table with an ICE table.
We take the weak acid, it is placed in water
it re-establishes an equilibrium with its conjugate base.
And when this donates to the water it will produce
H_3O+.
This is an ICE table
and we have to put molarity in there.
So we have to figure out the molarity
of the BH+. Well,
you get the molarity of BH+
we will need the mole of BH+
0.0010 moles
divided by the total volume.
Well, I choose to have
10 ml, and I need to write it as liters
which is 0.10 liters
of the acid. which is 0.10 liters
of the acid.
And that was my choice.
And I will have to add how many
liters
of the base were needed in order to get to the equivalence point.
Well if we look at our numbers here
I have half of the concentration
of the acid, that I have base. I have half of the concentration
of the acid, that I have base.
So I am only going to need half of that base.
So I am only going to need 5 ml.
But lets say, that didn't come to mind.
But you wanted to use the procedure we used before.
We look in our table and we see how much base there is.
We have 0.0010 moles of the base. We look in our table and we see how much base there is.
We have 0.0010 moles of the base.
And we say I don't want moles
I want volume, and the molarity of base
is .2 moles per liter.
And that tells me it is going to be is .2 moles per liter.
And that tells me it is going to be
equals 0.0050 liters, or 5 ml.
I place that there, so now I can know the molarity
of that solution.
So the concentration is 0.067 molar.
Now we can place that into our table down here at the bottom
because we need that value in order to fill in our ICE table.
0.067
don't care about that, these are zero.
Minus x, plus X, plus X
0.067, x, x.
And now this is a what? Is this an acid or a base?
It is an acid, so we are going to need the K_a
is equal to the B
times the H_3O+
divided the BH+
we plug in what we know. We don't know the K_a, but we know
K_b, so we use 1 x 10 ^-14
divided by
the K_b 4.4 x 10 ^-4
and that is going to be X squared [x^2]
over 0.067 - x. and that is going to be X squared [x^2]
over 0.067 - x.
And you will need to solve for X, so pause the video
put in your numbers and solve for X.
Did you come up with an X equal to
1.2 x 10 ^-6?
If so you did well.
That X is the
ICE table, that X
is the H_3O+ concentration.
So we take the negative log of the is the H_3O+ concentration.
So we take the negative log of the
value of x, we will get the pH.
And the pH equal 5.91 . value of x, we will get the pH.
And the pH equal 5.91 .
So there is our answer.
We check, is that below 7? Certainly it is it should
have been so that is a reasonable value.
For our pH.
And jsut look at this work.
Whenever you are at the equivalence point.
You will do and ICF table to figure out the
moles of the conjugate left in solution You will do and ICF table to figure out the
moles of the conjugate left in solution
and it will always be followed by at ICE table
so this is the lengthiest part
of a titration curve, is when you are at the equivalence point.' so this is the lengthiest part
of a titration curve, is when you are at the equivalence point.'
This problem is in place to show we are not
always calculating the pH
of a titration.
You learned about titrations
much earlier, not in the units here
of this class.
And this is a more typical titration problem.
You do a titration usually to
find something out about an unknown.
And Our unknown is an monoprotic acid
and I am going to call it HA, meaning it has one H on it. And Our unknown is an monoprotic acid
and I am going to call it HA, meaning it has one H on it.
And this monoprotic acid is being titrated with
sodium hydroxide a strong base. And this monoprotic acid is being titrated with
sodium hydroxide a strong base.
And we know that the proton swap
will apply and here is our reaction.
Now we are not wanting to know the pH so we don't
have to do all the work that we had to do before. But lets think about what we are trying to find out.
We are trying to find out the molar mass
of this substance.
And molar mass has units of grams per mole.
And they gave me the grams of this acid.
That number is easy enough to obtain.
Our real work is to obtain the moles of that acid.
Well, a titration
the purpose of the titration is to get to
the equivalence point so you can make a comparison between the moles of those substances.
They gave us, to obtain the moles of hydroxide
they gave us the molarity of the hydroxide
and the volume of the hydroxide.
And this will give us the moles of the hydroxide which is
1.16 x 10 ^ -3.
And I will put that in my
ICF table. 1.16 x 10 ^ -3.
Well when you
get to the equivalence point we know that we moles
of the HA are 1.16 x 10 ^ -3, that and equivalence point.
Now we don't have to do anything else here with this table.
Yes, it reacts until you produce the A minus
and all that, but what we really want to know is how much
HA was in there in terms of moles, so we can place it right here.
And now we have the grams
of acid, we have the moles of acid and we have a And now we have the grams
of acid, we have the moles of acid and we have a
molar mass of 110 grams per mole.
So this is the end of our learning objective 6
we are doing titrations between acids and bases where
one is weak and one is strong. we are doing titrations between acids and bases where
one is weak and one is strong.
This is very challenging for students it is really hard
because there is so many variation.
And you want to follow one model
and one model does not work.
So I have produced a whole series
of worked example problems for you to watch. So I have produced a whole series
of worked example problems for you to watch.
You need to be continually thinking about
why did she write the reaction she wrote
how do I know that is what I should write?
And if you can do that, and get it started
and examine what you have in the F line
and decide on your next course of action
you will be able to that working through these problems.
So work through them, watch me work through them
keep practicing, until you feel comfortable with writing the reaction
filling in your tables and obtaining your answers.