0:11
To do this you first have to choice the right
substances to be apart of that buffer.
You want to choice an acid whose
pK value is close to
the desired pH you are looking for.
Now why would that be?
We think of the Henderson-Hasselbalch equation
which applies to buffers.
0:34
The pK_a value
of the acid in that buffer
is going to be your starting point.
This term
how much base and how much acid you
put in there, that ratio base to acid
will tweak it up or down depending
upon whether you need it a little bit higher then that
pK_a value of pH or a little bit lower of that pK_a value.
But you want to start with
an acid whose pK_a value is
close to the desired pH.
If you are looking for a buffer that is
in the acid's range, meaning If you are looking for a buffer that is
in the acid's range, meaning
less than 7.
You are going to be going to
tables of acids, you are going to be looking at those acids
and finding one whose pK_a is close to what you want.
But if you wanted to make a buffer and finding one whose pK_a is close to what you want.
But if you wanted to make a buffer
that has a higher pH
lets say great then 7 in the
base range, then you want to be choosing
a starting material
which is a base.
So that you have got a basic buffer.
But again you are going to be looking at the K_a value
of its conjugate acid.
So if you were looking at
base like ammonia
you need to consider
acid ammonium
and say what is the K_a
value of ammonium
and gets its pK_a value.
So you are looking for something in that range.
Once you have decided
the substances that will make up your buffer
then you are going to take that Henderson-Hasselbalch equation
and you are going to solve for the ratio then you are going to take that Henderson-Hasselbalch equation
and you are going to solve for the ratio
of the base over the acid.
And that way you will know how much
of what concentration of those two items you are going to need.
In order to make that pH.
You might want also to consider the moles of base
over moles of acid, instead of
concentrations
Then once you know how many moles of each that you are going to need
or the ratios of moles you are going to need
then you could say well,
how many grams will i need to weight out
to get that many moles or
I want to do a dilution and I am going to figure out to get that many moles or
I want to do a dilution and I am going to figure out
molarity of the volume I need
to get that number of moles.
2:51
Consider the Henderson-Hasselbalch equation
pH equals pK_a Consider the Henderson-Hasselbalch equation
pH equals pK_a
plus the log pH equals pK_a
plus the log
and I am going to do it in term of moles. plus the log
and I am going to do it in term of moles.
Number of base over the number of moles of acid. and I am going to do it in term of moles.
Number of base over the number of moles of acid.
I will plug in that I am trying to Number of base over the number of moles of acid.
I will plug in that I am trying to
obtain a 4.80 pH
and I am going to use the negative log of the acid
which is 2.0 x 10 ^ -4
My goal will be to solve
for the log of
the number moles of base, over the number moles of acid.
4:22
If we choose that amount
then we will know that
if we multiply that by 12.6
that would give me the number of mole
in the base I would need. In other words I would need
1.26 moles of base.
So if I made a solution
that had those two amounts
.1 mole of acid
1.26 moles of base
then I would know I would have created a buffer
with a pH of
4.80 which is what I was trying to achieve.
It does not have to be
.1 moles. It does not have to be
.1 moles.
I choose any amount here .1 moles.
I choose any amount here
lets say I want to have more acid
than more base. Maybe I want to have
5:27
Now buffers
will quit working after a while.
You add an acid it will will quit working after a while.
You add an acid it will
drop the base portion of that
buffer down. If you keep on adding acid drop the base portion of that
buffer down. If you keep on adding acid
you keep dropping it down.
Buffering capacity is the amount of acid
of base that you can add to that buffer
without causing a large change of pH.
Eventually you wipe out that buffering capacity.
You will have used up the substances
and you no longer have a good ratio.
Of conjugate acid and conjugate base and you no longer have a good ratio.
Of conjugate acid and conjugate base
pairs in there, in your buffer.
If you look at this diagram down here in the bottom right.
We see that they both have the same
ratio of base to acid.
So the number of moles if we let each ratio of base to acid.
So the number of moles if we let each
one of those So the number of moles if we let each
one of those
figures rep[resent a mole
the number of moles of base to the number
moles of acid the number of moles of base to the number
moles of acid
is 4 to 4
or 1 over here.
The number moles of base
to the number of moles of acid is 2 to 2
it would have the same pH. to the number of moles of acid is 2 to 2
it would have the same pH.
And it would have, but one would have
better buffering capacity then the other.
6:32
So anytime you increase the concentration of the acid or
the base in your buffer, then you increase So anytime you increase the concentration of the acid or
the base in your buffer, then you increase
the capacity. the base in your buffer, then you increase
the capacity.
6:54
You could also do a ratio of
acid to base and i will maybe
draw it in like this.
Lets say I have got
four of the acid
and I only have two of its
conjugate base.
If I have a buffer like this
it has a good buffering capacity
to neutralize it has a good buffering capacity
to neutralize
7:22
Because there is quite a bit
of acid in that buffer
but it would not have as good a of acid in that buffer
but it would not have as good a
buffering capacity again but it would not have as good a
buffering capacity again
added acid
because the added acid would be neutralize
by the base form of that buffer.
So you could have a buffer
8:11
And so we are start off with this down here
and not all of this up there.
A less amount of this acid.
that this buffer would
be a good buffer against added acid.
Because there was quite a bit of the conjugate base in there.
but it would not be as affective again added
base because I am starting with a lesser amount.
Dr. Allison SoultLecturer
Dr. Kim WoodrumSr. Lecturer
