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4.04 Buffer: Preparation and Capacity

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Высшая химия

Кентуккийский университет

4.7 (оценок: 420) | Зарегистрировано учащихся: 50K

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.

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4.7 (оценок: 420)

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KO

Jul 17, 2017

Excellent content and delivery!\n\nThis course has many well-explained examples and practice questions to reinforce the understanding of the taught concepts. Thank you!

CW

Sep 17, 2015

I used this course as a significant portion of my preparation for the CLEP Chemistry exam, and passed today with an A! Thank you!

Из урока

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.

4.01 Buffers and the Common Ion Effect10:01

4.02 pH of Buffer Solutions14:04

4.02a Aqueous Equilibria Worked Example 13:58

4.03 Buffer Action25:21

4.03a Aqueous Equilibria Worked Example 25:22

4.03b Aqueous Equilibria Worked Example 35:11

4.04 Buffer: Preparation and Capacity8:45

4.05 Strong Acid - Strong Base Titration17:20

4.06 Titrations Involving Either a Weak Acid or a Weak Base36:45

4.06 Part 1 Aqueous Equilibria Worked Example3:46

4.06 Part 1.a - Aqueous Equilibria Worked Example4:06

4.06 Part 1.b - Aqueous Equilibria Worked Example4:55

4.06 Part 1.c - Aqueous Equilibria Worked Example4:21

4.06 Part 1.f Aqueous Equilibria Worked Example8:01

4.06 Part 1.h Aqueous Equilibria Worked Example3:57

4.06 Part 2.a Aqueous Equilibria Worked Example8:25

4.06 Part 2.b Aqueous Equilibria Worked Example5:53

4.07 Polyprotic Acid Titrations5:53

4.08 Indicators10:19

4.09 Solubility Equilibria4:50

4.10 Molar Solubility and the Solubility Product10:37

4.10.a Aqueous Equilibria Worked Example2:20

4.11 Molar Solubility and the Common Ion Effect7:20

4.11 2.b - Aqueous Equilibria Worked Example4:42

4.11 2.c Aqueous Equilibria Worked Example3:45

4.12 The Effect of pH on Solubility7:07

4.12.b Aqueous Equilibria Worked Example3:42

4.12c Aqueous Equilibria Worked Example3:30

4.13 Precipitation Reaction and Selective Precipitation15:03

4.13a Aqueous Equilibria Worked Example9:05

Преподаватели

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

Текст видео

We are ready for our next learning objective

and as it says here in the on screen we are going to be

talking buffering capacity

and how to prepare a buffer for a specific pH.

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.

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.

Now in this problem, I have already

told you what the substances are

that are going to be making that buffer.

I am going to show you the idea

of solving for the ratio.

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.

The negative log of 2.0 x 10 ^-4

is 3.70.

So I will subtract 3.70 from both

side and have 1.10.

Now we have to get rid of the log

log function

inverse log is 10 to that value.

So those will cancel

and we will be left with

12.6

equals the number moles of base

over the number of moles of acid.

Another way to write that would be

12.6 times the number of moles of acid

would give you the number of moles of base.

So if at this point you have all sort of

value that would work and give you a pH of that amount.

Lets say we will choose

a .10 moles

of the acid.

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

I want 1.0 moles of acid.

Well that would tell me that I would need

to have a solution with

12.6 moles of base in it.

But whatever, that ratio has got to be

12.6 to 1.

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.

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.

Mess up what is popping up next.

We have a diluted buffer and we have a concentrated buffer

and certainly the concentrated buffer

has a better buffering capacity.

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

an added base.

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

in this case because you needed a certain pH range.

That would be more effective in this case because you needed a certain pH range.

That would be more effective

against the addition of an acid verse the addition of a

base and vise versa.

This is just a diagram we have seen earlier and we

can imagine in this one

if we had

less acid to start with

then we had base.

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.

So that is the end of our learning objective for four

you know what buffer capacity is and you know how to

create a buffer with a specific pH range.

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