A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.
4.05 Strong Acid - Strong Base Titration
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From the lesson
Aqueous Equilibria
This unit continues and expands on the theme of equlibria. You will examine buffers, acid/base titrations and the equilibria of insoluble salts.
Taught By
Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer
We are ready to begin our next round of learning objectives
dealing with acid and base titration.
within these acid/base titrations
we are going to learn to calculate the pH
At various points
along the titration.
Now if you will think back to along the titration.
Now if you will think back to
what you have learning in the past
you have learned about acids and bases
titrations, most likely.
You might want to go back and review
what you have learned.
But one of the things you learned to do with
acid and base titrations is you are given
an acid
where you know the concentration and you are
trying to determine the concentration of
an unknown base.
You know the concentration
of a base and you are trying to determine the
pH or the concentration of an unknown acid.
And its a stoichiometry problem where you are
covering from the molarity and the volume of
the known or the standard solution
getting into moles of that
substance and determining
the moles of the other substance.
Within these units
that we are talking about here Within these units
that we are talking about here
and the next couple of learning objectives
we are going to look at how pH
changing during the course of this.
We are going to look at three types of titration.
We are going to begin
in this lesson with strong acids
and strong base titrations.
This will be learning objective number 5.
In learning objective number 6
we are going to have one of those
be weak. Either the acid will be weak
or the base will be weak. be weak. Either the acid will be weak
or the base will be weak.
We will never do calculations in which
both the acid and the base are weak.
There are very many equilibria involved
and it gets very complicated.
So we are going to do make sure that at least and it gets very complicated.
So we are going to do make sure that at least
one of those two will be strong.
So lets being, our learning objective number
5 will be
calculating the pH of a strong acid
and strong base titration
and any point along
the titration, from the beginning
to the, what we call the equivalence point
and beyond.
What happens when a strong acid and a strong base
react? We we can write the reaction
as follows.
Here is an example of a strong acid
and a strong base. The
cation of the base and the anion of the
acid will form the salt.
And the H of the acid
and the OH of the base
will produce the water.
We are going to consider just the net
ionic equation.
It would be H+ of the acid
and OH- of the base
combining to form water.
Or we could and I am going to
back this up, we could write this
typically will write this
with H_3O+
for the strong acid with H_3O+
for the strong acid
and OH- for the strong base.
And then the reaction is just a little bit different
we have two waters.
Because remember when a strong acid
dissolves, you could either say Because remember when a strong acid
dissolves, you could either say
it 100%
ionizes
to produce those ions.
Or you could say, it is to produce those ions.
Or you could say, it is
reacting with water
to 100%
give you the anion
NaH_3O+
so either way, and I typically choose
writing strong acids as H+.
Now this is a one-way reaction.
And with a one-way reaction we are going to use
what is called an ICF table
and we saw those tables
we when were dealing with adding and acid
to a buffer or a base to a buffer.
We know that it is a one-way
push it to completion type of reaction.
So lets look at how pH is changing
during the course of a titration
prior to actually getting into the calculations.
Here is a schematic of the reaction taking place.
We have in the flask a strong acid. Here is a schematic of the reaction taking place.
We have in the flask a strong acid.
That is our .1 molar HCl.
And lets say we have 25 ml of it.
We are going to start adding base
to this and I have got a sodium hydroxide in the
portion I am interested in is the OH- of that base.
And here is our reaction H+ and OH- forming water.
I am going to plot, and kind of determine
pH at every point along this titration
as the base is being added.
At the very beginning
before I have added any base what so ever
We have got the pH of this strong
acid. Well, if it is .1 molar We have got the pH of this strong
acid. Well, if it is .1 molar
and we take the negative log
of .1
we will have a pH of 1.
So we have a red dot there, if you look at the graph
at one, the beginning.
Now we are add a little bit
the base, and we are going start adding that base Now we are add a little bit
the base, and we are going start adding that base
into the flask.
When a little bit, look at the base up here into the flask.
When a little bit, look at the base up here
when a little bit of the base is added
it will react with some
of the acid.
And the base will go away, the base will be turned into water.
But there will be a lot of acid left.
But not as much as we started with.
So the concentration
of that acid is
going to be dropping a little bit. of that acid is
going to be dropping a little bit.
during the course of this titration.
Well, if that concentration is
dropping a little bit the pH
is going to come up a little bit so we see
a slight rise of
the pH. Well, we keep adding and
we keep adding and it going to rise and rise and rise.
But it is not changing drastically just
slowly increasing as the
H+ concentration
is dropping a little at a time
as we continue to add a little bit
base and add a little bit of base.
But there will come a time where
the amount of acid and the amount of
base are exact equal the amount of acid and the amount of
base are exact equal
and we have added just enough OH-
to react with all of the H+.
At that point we have reached what is called the equivalence point.
There number of moles of base is going to be stoichiometrically
equal the number of moles of acid.
So there are a certain number of moles of acid in here
right, we had a certain number of moles and we
kept on adding until all of those right, we had a certain number of moles and we
kept on adding until all of those
we finally converted to acid
and we had the picture
and all the acid had been converted
to water, and we have the right on the right.
When the titration is between
a strong acid and a strong base When the titration is between
a strong acid and a strong base
you will have converted all the H+
and OH- over to water.
And the pH of water is 7. and OH- over to water.
And the pH of water is 7.
So always the pH is going to be 7
at the equivalence point when they are both strong.
But we can keep on adding
we can go beyond
the equivalence point
and what happens when we continue to add
base? If we were take this system
and add some more
base to there, suddenly we got
base present.
We already have consumed all of the acid. base present.
We already have consumed all of the acid.
There is no acid left to neutralize.
so the pH is going to shoot up.
It is going to shoot up very rapidly and then just
gradually increase as we add more and more
base to the flask.
So we are going to be able to calculate the pH
at any point. So lets go back to this procedure.
This is a similar procedure at any point. So lets go back to this procedure.
This is a similar procedure
that we saw when we added
an acid or base to a buffer.
The first thing we will have is a one way reaction
in between the acid and the base.
We will use an ICF table
and put moles into the table which is a way
of doing a limiting reactant problem.
We will examine that final line
and determine our next course of action.
At this point there could be three
possibilities of what could happen.
It there remains a conjugate base pair
then we would use the Henderson-Hasselbalch equation.
This would never happen
when we are in this portion of our
lesson, because we are dealing with
strong acid and strong base. There will never be a weak
thing present and its conjugate.
But if there is a titration and it is weak
then this would occur prior to the equivalence point and we will
come back to that later.
If there remains
only a weak acid or a weak base present
this occurs when we are at the equivalence point
if one of them is weak. Again,
in this portion of our lesson this will not happen.
What will happen in this portion of the lesson
is there may remains
some strong acid, or some
strong base, and we put the word strong here as well. some strong acid, or some
strong base, and we put the word strong here as well.
And when that happens you would
calculate the pH directly
from the molarity of that
strong acid or strong base that is left over.
With this procedure
we are going to hold and ready
these two portions
there are not going to happen when they are both strong.
But we will see them later
when we are in learning objective
number 6.
OK we are going to determine the pH
beginning with a small amount
of sodium hydroxide being added to the acid.
So we are only adding 5 ml.
Begin by writing the reaction.
The strong acid I
use H_3O+ to represent the strong acid.
I let OH- represent the strong base.
I will form 2H_2O.
I will do an ICF table according to my instructions. I will form 2H_2O.
I will do an ICF table according to my instructions.
One way go to completion.
Now I need the moles of H_3O+
and the moles of OH-. Now I need the moles of H_3O+
and the moles of OH-.
Lets begin with the moles of H_3O+
coming from the acid.
The molarity of HCl coming from the acid.
The molarity of HCl
is .1.
If I multiple by the volume which is
.025 liters
that will produce from me .0025
moles of H_3O+.
and I can put that into the table.
For the moles of base
I see that the sodium hydroxide has one OH in it.
So the moles of OH would be the molarity I see that the sodium hydroxide has one OH in it.
So the moles of OH would be the molarity
of the sodium hydroxide
times its volume, again in liters
0.0050 liters.
That is going to give me 0.005
moles of base. And I can plug that into my table.
I don't care about the water, I am just going to put a line through that.
I am going to consume the smaller of the too.
That is the base. I am going to consume the smaller of the too.
That is the base.
I would have, of course, produce
some of it. I had a lot
to begin with a few more moles
that is still going to be a lot there
and I will be left with that is still going to be a lot there
and I will be left with
none of the base.
So at this point, I have
H_3O+
in solution, acidic solution, and it is a
strong acid. So I can obtain the pH
by taking the negative log
of the H_3O+ concentration. by taking the negative log
of the H_3O+ concentration.
Well, this is not
the H_3O+ concentration.
I plugged everything in this table using moles.
So this is the moles of H_3O+.
So if I want to know the concentration of H_3O+
I have to divide by the volume. So if I want to know the concentration of H_3O+
I have to divide by the volume.
And I want to divide by the total volume.
So I need to take the negative log
of the 0.002 So I need to take the negative log
of the 0.002
moles and I have
25 ml of the acid
5 ml of the base
so that is .030 5 ml of the base
so that is .030
liter, 30 ml, or .030 liters.
When I
divide those numbers out
it is equal to 0.067.
That is the concentration
when I take the negative log of that That is the concentration
when I take the negative log of that
value is 1.18.
So that is the pH of the solution.
So it should be acidic.
We are before the equivalence point
we neutralize all the base We are before the equivalence point
we neutralize all the base
and we have converted the base
to water but there is still a lot of extra acid in there.
Now lets get to the equivalence point.
We are going to add 25 ml
of the base to the 25 ml of
the acid. Since there are both at the same
molarity, we know we are at the equivalence point.
But lets imagine that you didn't realize that.
And you start redoing the problem But lets imagine that you didn't realize that.
And you start redoing the problem
blindly as we have been before.
Eventually you are going to recognize you are at the equivalence point.
We would take the molarity and the volume
of the acid. It is the same molarity
of the same volume of acid we had before.
So we are going to place that number
.0025 here. So we are going to place that number
.0025 here.
And if wanted to figure out the moles of
hydroxide we do the same thing.
We take the molarity
times the volume in liters
and we see that is it .0025 moles.
So at this point you might recognize
that hey we are at the equivalence point.
But lets say you didn't at this point.
This is what I see some students doing.
They just keep on working through the problem and they have a zero This is what I see some students doing.
They just keep on working through the problem and they have a zero
here and a zero here. They just keep on working through the problem and they have a zero
here and a zero here.
And then they scratch their heads and say
I have nothing to base my calculation on.
Keep in mind that if you neutralize all your acid and base
you will only have water and what is the pH of water? It is 7.
So at some point you are going to recognize you are at the equivalence point.
And whenever you are at the equivalence point between a
strong acid and a strong base
the pH is 7. strong acid and a strong base
the pH is 7.
OK now I am going to have you
stop the video after a bit
and try to work through this problem. stop the video after a bit
and try to work through this problem.
We have got a titration between
a strong acid and a strong base. The same one. We have got a titration between
a strong acid and a strong base. The same one.
But now we have gone beyond the equivalence point
and added excess sodium hydroxide.
So I want to see if you can calculate the pH of this solution.
Well, if you came up with 11.96, very well done.
That is the pH of that solution.
If you struggled with that
I am going to go through how to work
and get that 11.96.
You might not need to see this
in which case you can of course skip ahead.
So we have our reaction that we have been working with.
H_3O+ plus OH-
producing 2 waters.
We have our ICF table
the amount of acid hasn't changed
in any one of these problems. the amount of acid hasn't changed
in any one of these problems.
But now the amount of in any one of these problems.
But now the amount of
base has certainly changed.
30 ml of .1 molar base has certainly changed.
30 ml of .1 molar
is 0.0030 .
So we have more of the
base then we have of the acid.
So know when we consume
smaller amount it is the So know when we consume
smaller amount it is the
acid that going away.
We are left with 0.005 moles
of the base.
Since it is a base that is left over we can
first calculate
the pOH.
That would be equal to the negative log
of the hydroxide concentration.
Well this is just moles so we have of the hydroxide concentration.
Well this is just moles so we have
to divide by the total volume.
The total volume would be
25 ml of the acid The total volume would be
25 ml of the acid
and 30 ml of the base thats
55 ml
or .055 liters.
That would be the negative log
of 0.009.
That gives me a pOH
of 2.04.
Of course the pH
is simply 14 minus that 2.04.
That is how we get the 11.96. is simply 14 minus that 2.04.
That is how we get the 11.96.
Now hopefully when you saw this
problem, you realized it had to be basic
and you eliminated these two answers right off the bat.
But you could not distinguish between the other two.
Without first doing this process.
So that is the end of
learning objective 5 in which
we determined the pH
when the acid and the base are strong. we determined the pH
when the acid and the base are strong.
We have
a procedure to follow. We do an ICF
table there will always be either at the a procedure to follow. We do an ICF
table there will always be either at the
equivalence point in which case the pH is 7
no calculation is needed.
Or you have got more acid
strong acid in there, or more strong base in there
and you can calculate of either the acid or the base
using the negative log of that concentration.
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