LESSON PLAN

in
General Chemistry 2

School : Bestlink College of the Philippines

Teacher : Mr. John Carlo Mateo Cerico Grade Level and Quarter : Grade 7 Quarter 1

Date: June 2022 Learning Area : Virtual Classroom (GoogleMeet)

Time : 7:00 – 7:30 AM Topic : Buffer Effectiveness

I. OBJECTIVES:

At the end of this module student should be able to:

1.1 Compute the proportional quantities of a weak acid and its conjugate base
required to produce a buffer solution with the specified pH
1.2 Ascertain the buffering capability of various solutions; and
1.3 Recognize the significance of various buffers in natural systems

II. SUBJECT MATTER and REFERENCE

2.1 General Chemistry 2, Module 17, Page 1

2.2
https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:
acids-and-bases/x2eef969c74e0d802:buffers/v/properties-of-buffers

2.3https://www.jove.com/science-

education/11412/buffer-

effectiveness

2.4 Concepts:

 Relative amounts of Acid and Base - The pH of a buffer depends on the ratio
[base]/[acid] rather than on the particular concentration of a specific solution.
 Identifying Acid and Conjugate Base Pairs - A buffer is an aqueous solution
consisting of a mixture of a weak acid and its conjugate base or a weak base and
its conjugate acid. Therefore, it is very important to be able to identify acid and
conjugate base pairs.
 Buffer Range and Capacity - A buffer’s capacity is the pH range where it works as
an effective buffer, preventing large changes in pH upon addition of an acid or
base.

III.INSTRUCTIONAL MATERIALS

Powerpoint, Paper and Pen, Calculator

IV. PROCEDURE
 Teacher’s Activities Student’s Activities
INTRODUCTION/ PRELIMINARY ACTIVITIES

Greetings and Prayer: “Good Morning, Sir!”

“Good morning, class.” (One student will lead)
“Let us pray first. Lead us in prayer...’

Checking Attendance:
(listed names in the paper after the activity)

Checking Assignment

Review of Previous Lesson “The Brønsted-Lowry theory describes acid-

“What is Bronsted-Lowry Theory all base interactions in terms of proton transfer
about?” between chemical species. ”
MOTIVATION
“Guess and Tell!”

INSTRUCTION: Look at the jumbled letters and refer to the pictures above it then try to guess
what those letters and pictures represent. After guessing the correct answer, please explain
what comes to your mind when you hear those words.

SIDAC ESBAS REFFUB NOITARTNECNOC

LESSON PROPER
Relative Amounts of Acid and Base:

The pH of a buffer depends on the ratio

[base]/[acid] rather than on the particular
concentration of a specific solution. A buffer is an
aqueous solution consisting of a mixture of a
weak acid and its conjugate base or a weak base
and its conjugate acid. A buffer’s pH changes very
little when a small amount of strong acid or base
is added to it. It is therefore used to prevent
change in the pH of a solution upon addition of
another acid or base. The pH of a buffer depends
on the ratio [base]/[acid] rather than on the
particular concentration of a specific solution.
The exact ratio of the base to the acid for a
desired pH can be determined from the Ka value
and the Henderson-Hasselbalch equation.

Questions:

Who developed the Henderson-Hasselbalch “Sir, that would be no other than Lawrence
equation? Joseph Henderson”

Alright! That is correct nak!

(Explain brief information about Lawrence Joseph
Henderson)

Who re-expressed Mister Henderson’s formula in

logarithmic term? “Sir, His name is Karl Albert Hasselbalch”

That is also correct! Thank you so much for

that! You did well class for doing an advance
reading just like what I told you to do for
today’s lesson.

Karl Albert Hasselbalch later re-expressed that

formula in logarithmic terms, resulting in the
Henderson-Hasselbalch equation [1].
Hasselbalch was using the formula to
study metabolic acidosis, which results from
carbonic acid in the blood.

Just like what Sir Cerico said, One way to

determine the pH of a buffer is by using the
Henderson–Hasselbalch equation, which is pH
= pKₐ + log([A⁻]/[HA]). In this equation, [HA]
and [A⁻] refer to the equilibrium
concentrations of the conjugate acid–base
pair used to create the buffer solution. When
[HA] = [A⁻], the solution pH is equal to the
pKₐ of the acid.

Does the class understood that well?

“Yes we did Sir”

Very well, since you said that we can move on now
to the Henderson-Hasselbalch equation.

Are you all ready?

Alright, “Yes we are Sir!”

In chemistry, the Henderson-
Hasselbalch (frequently misspelled Henderson-
Hasselbach) equation describes the derivation
of pH as a measure of acidity (using pKa, the acid
dissociation constant) in biological and chemical
systems.
The equation is also useful for estimating the pH of
a buffer solution and finding the equilibrium pH
in acid-base reactions (it is widely used to
calculate isoelectric point of the proteins).

The Henderson-Hasselbalch equation is an

equation that's often used to calculate the pH of
buffer solutions. Buffers consists of a weak acid
and its conjugate base.

So for a generic weak acid, we could call that HA,

and therefore, its conjugate base would be A-.

“HA/A -”
To calculate the pH of the buffer solution, we
would find the pKa of the weak acid, and to that
we would add the log of the concentration of the
conjugate base divided by the concentration of the
weak acid.

“pH = pKₐ + log([A⁻]/[HA])”

Let's use the Henderson-Hasselbalch equation to

calculate the pH of an aqueous buffer solution that
consists of acetic acid and its conjugate base, the
acetate anion.

= CH3cooH

= CH3Coo-

And let's use this particulate diagram to help us

calculate the pH of the buffer solution. Remember
that the goal of a particulate diagram is not to
represent every particle in the solution, but to give
us an idea about what's going on in the entire
solution. And also, when looking at the particulate
diagrams of buffer solutions, water molecules and
cations are often left out for clarity.

Let's count the number of particles of acetic acid in

our particulate diagram. So in our diagram, there
are five particles of acetic acid,

= 5 (CH3CooH)

and for the acetate anion, there are also five.

= 5 (CH3Coo-)

Because there are five particles of both acetic acid

and the acetate anion, the concentration of acetic
acid is equal to the concentration of the acetate
anion.

[CH3CooH = CH3Coo-]

Next, let's think about the Henderson-Hasselbalch

equation. Our goal is to calculate the pH of this
buffer solution represented in the particulate
diagram.

pH = pKₐ + log([A⁻]/[HA])
 Ka = 1.8 x 10-5
Ka = -log Ka
Ka = -log (1.8 x 10-5) = 4.74

And so first, we need to know the pKa of the weak

acid, which is acetic acid.
At 25 degrees Celsius, the Ka value for acetic acid
is equal to 1.8 times 10 to the negative fifth. The
Ka value is less than one because acetic acid is a
weak acid. To find the pKa of acetic acid, we take
the negative log of the Ka value. So the negative
log of 1.8 times 10 to the negative fifth is equal to
4.74.

So we can go back to the Henderson-Hasselbalch

equation

pH = pKₐ + log([A⁻]/[HA])

and write that the pH is equal to the pKa, which

we just calculated to be 4.74 plus the log of the
concentration of the conjugate base. And the
conjugate base is the acetate anions, so let's write
that in here, CH3COO-, and that's divided by the
concentration of the weak acid, which is acetic
acid, CH3COOH.
[CH3Coo-]
pH = 4.74 + log
[CH3CooH]
We've already figured out that the concentration
of acetic acid is equal to the concentration of the
acetate anion. Therefore, the concentration of the
acetate anion divided by the concentration of
acetic acid is just equal to one. And the log of one
is equal to zero. So let's go ahead and write that in
here, the log of one is equal to zero. Therefore,
the pH of the buffer solution is equal to 4.74 plus
zero or just 4.74. 

pH = 4.74 + 0
or
pH = 4.74

So whenever the concentration of the weak acid is

equal to the concentration of the conjugate base,
the pH of the buffer solution is equal to the pKa of
the weak acid.

[HA] = [A-] pH = pKa

Let's look at another particulate diagram. We still

have an acetic acid-acetate buffer solution.
However, this is a different buffer solution than
the previous problem. So let's count our particles.
 = 6 (CH 3CooH)

= 4 (CH 3Coo-)

[CH3CooH] > [CH3Coo-]

For acetic acid, there are six particles and for the
acetate anion, there are only four. Since we have
more acetic acid particles than acetate particles,
the concentration of acetic acid is greater than the
concentration of the acetate anion.

We can use the Henderson-Hasselbalch equation

to think about the pH of this buffer solution.

pH = pKₐ + log([A⁻]/[HA])

So the pH is equal to the pKa, which we calculated

in the previous problem for acetic acid, it's 4.74 at
25 degrees Celsius, plus the log of the
concentration of the acetate anion, divided by the
concentration of acetic acid.
[CH3Coo-]
pH = 4.74 + log
[CH3CooH]
In this case, the concentration of acetic acid is
greater than the concentration of the acetate
anion. Therefore, we have a smaller concentration
divided by a larger concentration. So we have a
number less than one. And the log of a number
less than one is negative. Therefore, all of this
would be negative or less than zero. So we would
be subtracting a number from 4.74.

Therefore, we can say the pH of the solution

would be less than 4.74. 

pH < 4.74

Let's do one more particulate diagram of an acetic

acid-acetate buffer solution. Once again, we count
our particles.

= 4 (CH 3CooH)

= 6 (CH3Coo-)
 [CH3CooH] < [CH3Coo-]

So for acetic acid, this time, there are four

particles and for the acetate anion, this time, there
are six particles. Since we have only four particles
of acetic acid and six particles of the acetate anion,
the concentration of acetic acid is less than the
concentration of the acetate anion or we could say
the concentration of the acetate anion is greater
than the concentration of acetic acid.

pH = pKₐ + log([A⁻]/[HA])

pH = 4.74

pH > 4.74

Thinking about the Henderson-Hasselbalch

equation, once again, the pKa is equal to 4.74, and
we need to think about the ratio of the
concentration of the acetate anion to the
concentration of acetic acid. For this example, the
concentration of the acetate anion is greater than
the concentration of acetic acid. Therefore, the
ratio would be greater than one, and the log of a
number greater than one is positive or greater
than zero. Therefore, we would be adding a
number to 4.74. So for this buffer solution, the pH
would be greater than 4.74.

So Class, did you understand how to compute

Henderson-Hasselbalch equation?

If that’s the case, can anyone tell or summarize

what we've learned from our three different
particulate diagrams? Let’s go first to the first “Yes Sir!”
diagram.

Alright nak, Go ahead

Very good nak! That was a very good answer! “Me Sir! I can answer that question of yours”
Thank you so much for that.

Just like what your classmate stated, In the first

example, the concentration of the weak acid was “*The student answer the question*”
equal to the concentration of the conjugate base.
And therefore, the pH of the buffer solution was
equal to the pKa of the weak acid.

Do we understand what the first diagram

represents?

How about the second diagram?

Very good! Thank you so much for that nak! You “Yes Sir!”
all did well.
So to make it simpler, in the second example, the
concentration of the weak acid was greater than
the concentration of the conjugate base. And “*The student/s answer the question*”
therefore, the pH of the buffer solution is less than
the pKa of the weak acid.

Next, how about the third diagram?

Thank you so much for that nak! Very good

answer!

And in the third example, the concentration of the “*The student/s answer the question*”
weak acid was less than the concentration of the
conjugate base. Therefore, the pH of the buffer
solution was greater than the pKa of the weak
acid. 

I can see that everyone here today did what I told

you to do last meeting. Very good class!

So always remember that if we know the pH of a

buffer solution, we can think about the
Henderson-Hasselbalch equation to think about
the relative concentrations of the weak acid and
the conjugate base.

For example, if we have a particular buffer solution

and we know the pH of the buffer solution is less
than the pKa of the weak acid, we know that in
that buffer at that moment in time, the
concentration of the weak acid is greater than the
concentration of the conjugate base.

Identifying Acid and Conjugate Bases:

A buffer is an aqueous solution consisting of a

mixture of a weak acid and its conjugate base or a
weak base and its conjugate acid. Therefore, it is
very important to be able to identify acid and
conjugate base pairs. The conjugate acid is created
by accepting (adding) a proton (H+) donated by
the conjugate base.

A buffer’s pH changes very little when a small

amount of strong acid or base is added to it.
Therefore, it can be used to prevent change in the
pH of a solution. Buffer solutions are used as a
means of keeping pH at a nearly constant value in
a wide variety of situations.

One of the main requirements of a buffer is that it

have the capacity to control pH after the addition
of a reasonable amount of acid or base. In other
words, there must be a large-enough
concentration of acetic acid in an acetic
acid/acetate ion buffer, for example, to consume
all of the hydroxide ions that may be added.

Did you understand what is conjugate acid and

conjugate bases are?

If that’s the case then, How can you tell an acid

from a conjugate base?

“Yes Sir!”
Yes! That is right. A conjugate acid contains one
more H atom and one more + charge than the
base that formed it. A conjugate base contains one “*The student/s answer the question*”
less H atom and one more - charge than the acid
that formed it.

Thank you so much for that nak. Very good

Now, we can move forward. Without further ado,
let’s do futher!

May I present, the buffer Range and Capacity!

A buffer’s capacity is the pH range where it works

as an effective buffer, preventing large changes in
pH upon addition of an acid or base.

A buffer solution usually contains a weak acid and

its conjugate base. When H+ is added to a buffer,
the weak acid’s conjugate base will accept a
proton (H+ ), thereby “absorbing” the H+ before
the pH of the solution lowers significantly.

Similarly, when OH– is added, the weak acid will

donate a proton (H+ ) to its conjugate base,
thereby resisting any increase in pH before shifting
to a new equilibrium point. In biological systems,
buffers prevent the fluctuation of pH via processes
that produce acid or base by-products to maintain
an optimal pH.

Now to make it more understandable, let’s discuss

the two.

A buffer has an effective pH range of one pH unit

on either side of the pKₐ value for the weak acid. If
the pH of a buffer goes out of this range, the
buffer will no longer be effective at resisting large
changes in pH.

Buffers consists of a significant amount of a weak

acid, which we will represent as HA and the
conjugate base to the weak acid, which we will
represent as A-. Buffer solutions resist large
changes in pH.

However, buffers are only effective over a certain

range of pH values. We are going to use the
Henderson-Hasselbalch equation to find the
effective pH range of a buffer. 

pH = pKₐ + log([A⁻]/[HA])

Looking at the Henderson-Hasselbalch equation,

the pH of the buffer solution is equal to the pKa of
the weak acid, which would be HA, plus the log of
the concentration of the conjugate base divided by
the concentration of the weak acid. And it's this
ratio of the concentration of the conjugate base to
the concentration of the weak acid that
determines if a buffer is effective or not.
 [HA] = [A -]

Buffer solutions are most effective at resisting a

change in pH in either direction when the
concentration of the weak acid is equal to the
concentration of the conjugate base. 

pH = pKₐ + log([A⁻]/[HA]) } 0

pH = pKₐ + 0

And when the concentrations are equal to each

other, the ratio is equal to one, and the log of one
is equal to zero. Therefore, when the
concentrations are equal to each other, the pH of
the buffer solution is equal to the pKa of the weak
acid plus zero. So we could just say that the pH is
equal to the pKa when the concentration of the
weak acid is equal to the concentration of the
conjugate base.

Again, the hasselbalch equation allows us to find

the outermost points of effective range

Concentration of the conjugate base is 10 times

the concentration of the weak acid

pH = pKₐ + log([A⁻]/[HA])

pH = pKₐ + log 10/1

pH = pKₐ + 1

Concentration of weak acid is 10 times the

concentration of the conjugate base.

pH = pKₐ + log([A⁻]/[HA])

pH = pKₐ + log 1/10

pH = pKₐ - 1

EFFECTIVE PH RANGE:
pH = pKₐ +- 1
BUFFER CAPACITY
The amount of acid or base that you can add to a
buffer without causing a large change in the pH
absolute concentrations of buffer components ↑,
buffer capacity ↑, the more concentrated the
weak acid and conjugate base are that compose
the buffer, the higher the capacity is relative
concentrations of buffer components become
more similar to each other

GENERALIZATION
INSTRUCTION: Write the summary of your
reflection on what you have learned in this module
not less than 150 words
APPLICATION
INSTRUCTION: Write TRUE if the statement is
correct and FALSE if the statement is wrong.

_____________ 1. A buffer is most effective when

the concentrations of acid and conjugate base are
approximately equal.
_____________ 2. A buffer is very important to be
able to identify acid and conjugate base pairs.
_____________ 3. A buffer has the capacity to
control the pH after the addition of a reasonable
amount of acid or base.
_____________ 4. A buffer’s capacity is the pH
range where it works as an effective buffer.
_____________ 5. Buffer cannot prevent large
changes in Ph
_____________ 6. When H+ is added to a buffer,
the weak acid’s conjugate base will donate a
proton (H+) to its conjugate base
_____________ 7. When OH– is added, the weak
acid will accept a proton (H+) to its conjugate base
_____________ 8. A titration curve visually
demonstrates buffer capacity.
_____________ 9. The addition of base or acid
greatly affects the pH of the solution drastically.
_____________ 10. The exact ratio of the base to
the acid for a desired pH can be determined from
the Ka value and the Henderson-Hasselbalch
equation

ACTIVITY 2
INSTRUCTION: Match column A with the
corresponding conjugate bases on column B.
Column A Column B
______1. Strong acid A. Strong conjugate base
______2. Strong base B. Weak conjugate base
______3. Weak acid C. It depends on the ratio
[base]/[acid
______4. Weak base D. Strong conjugate acid
______5. pH of the
buffer E. Weak conjugate acid

EVALUATION

Direction: choose the correct answer to the

following questions.

1. An effective buffer neutralizes _____ to

______ amount of added acid and base.
a. Small to large
b. Large to moderate
c. Small to moderate
d. High to moderate
2. RELATIVE AMOUNT OF ACID AND BASE?
a. Buffer are most effective when the
concentrations of the acid and its
conjugate base are equal.
b. Buffer are most effective when the
concentrations of the acid and its
conjugate base are unequal
 c. Buffer are most effective when the
concentrations of the acid is higher than
the conjugate base
d. Buffer are most effective when the
concentrations of the acid and its
conjugate base are high
3. ABSOLUTE CONCENTRATIONS OF ACID
AND CONJUGATE BASE?
a. Buffer are most effective when the
concentrations of the acid and its
conjugate base are equal.
b. Buffer are most effective when the
concentrations of the acid and its
conjugate base are unequal
c. Buffer are most effective when the
concentrations of the acid is higher than
the conjugate base
d. Buffer are most effective when the
concentrations of the acid and its
conjugate base are high.
4. The buffer only effective if they keep
within ____ unit of change
a. 2 pH
b. 1 pH
c. 4 pH
d. 12 pH
5. How much added acid or base to the
buffer can effectively neutralize
a. Buffer solution
b. Buffer range
c. Buffer effectivity
d. Buffer capacity
6. The pH range over which a particular acid
and its conjugate base can be effective
a. Buffer solution
b. Buffer range
c. Buffer effectivity
d. Buffer capacity
7-10 write the Halssenbalch equation

Key to correction:
1.C
2. A
3. D
4. B
5.D
6.C
7-10: pH=pKa+log(HA/A-)

ASSIGNMENT
Review for your Final Exam good luck
class!

CLOSING REMARKS
"Before we end our class, let's have
now a closing prayer."
“Goodbye class!”
 A student will lead the closing prayer.

“Goodbye Sir and thank you”

Prepared by: Submitted to:

John Carlo M. Cerico Engr,Joel R. Cajipe, FRIEDr

Student Practice Teacher SPT Coordinator