This document discusses buffers and the Henderson-Hasselbalch equation for calculating buffer pH. It defines buffers as solutions containing approximately equal concentrations of a weak acid and its conjugate base, or a weak base and its conjugate acid. The Henderson-Hasselbalch equation relates the pH of a buffer to its pKa value and the concentration ratio of the conjugate base to weak acid. The document provides an example calculation of the pH of an acidic buffer initially and after addition of hydrochloric acid, demonstrating how buffers resist changes in pH from small additions of acid or base.

1. Acid-Base Equilibria (Pt. 10)
Buffers and the Henderson-
Hasselbalch Equation
By Shawn P. Shields, Ph.D.
This work is licensed by Dr. Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0
International License.

2. What is a Buffer?
A buffer is a solution of approximately
equal concentrations of either
A weak acid and its conjugate base
or
A weak base and its conjugate acid

3. Acidic and Basic Buffers?
Buffers made from a weak acid and its
conjugate base are called “acidic
buffers.”
A buffers made from a weak base and
its conjugate acid are called “basic
buffers.”

4. Buffers
Buffers resist changes in pH
due to small additions of acid
or base.

5. Henderson-Hasselbalch Equation
Calculating the pH of a buffer solution
𝐇𝐀 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+
𝐚𝐪 + 𝐀−
(𝐚𝐪)
pH = pKa + log
A−
HA
conjugate
base
weak acid

6. Henderson-Hasselbalch Equation
Use this form of the equation for acidic
and basic buffers.
Always calculate pKa for the acid.
pH = pKa + log
A−
HA
base
acid

7. Example: Calculating the pH of a Buffer
Calculate the pH of a buffer
made from 0.28 M HNO2 and
0.23 M NO2
.
The Ka for HNO2 is 4.6  10-4.

8. Example: Calculating the pH of a Buffer
Calculate the pH of a buffer made from 0.28 M
HNO2 and 0.23 M NO2
. The Ka for HNO2 is
4.6  10-4.
First, calculate pKa for the acid.
pKa =  log Ka =  log (4.6  10-4) = 3.34

9. Example: Calculating the pH of a Buffer
Calculate the pH of a buffer made from 0.28 M HNO2
and 0.23 M NO2
. The Ka for HNO2 is 4.6  10-4.
Plug in pKa and the concentrations of the
acid and conj base.
pH = pKa + log
A−
HA
pH = 3.34 + log
0.23
0.28
= 3.25
base
acid

10. Example: Calculating the pH of a Buffer
after the Addition of Acid or Base
A buffer is prepared using 0.28 M HNO2
and 0.23 M NO2
. The pKa for HNO2 is
3.34.
What is the pH of this buffer after
the addition of 0.05 M HCl?

11. Example: Calculating the pH of a Buffer
after the Addition of Acid or Base
A buffer is prepared using 0.28 M HNO2 and 0.23 M NO2
.
The pKa for HNO2 is 3.34. What is the pH of this buffer
after the addition of 0.05 M HCl?
HCl is a strong acid.
It will react completely (and immediately)
with the conjugate base (NO2
) to form weak
acid (HNO2).

12. Example: Calculating the pH of a Buffer
after the Addition of Acid or Base
A buffer is prepared using 0.28 M HNO2 and 0.23 M NO2
.
The pKa for HNO2 is 3.34. What is the pH of this buffer
after the addition of 0.05 M HCl?
Since we added 0.05 M HCl…
0.05 M of the conjugate base (NO2
) will
react (be used) to form an additional 0.05 M
weak acid (HNO2).

13. Example: Calculating the pH of a Buffer
A buffer is prepared using 0.28 M HNO2 and 0.23 M
NO2
. The pKa for HNO2 is 3.34. What is the pH of
this buffer after the addition of 0.05 M HCl?
Plug in pKa and the NEW concentrations of
the acid and conj base.
pH = 3.34 + log
0.23 − 0.05
0.28 + 0.05
= 3.08
0.18
0.33

14. Buffers Resist pH Changes Due to
Addition of Acid or Base
The pH of the buffer did not change much due to
the addition of 0.05 M HCl.
pH = 3.25 versus pH = 3.08 after addition.
The buffer resisted the pH change
due to the addition of acid.

15. Compare: Calculating the pH of a 0.05 M
Strong Acid Solution
What is the pH of water (not a buffer) after the
addition of 0.05 M HCl
pH =  log (0.05) = 1.30
The pH changed due to the addition of
acid a lot (from pH = 7 to 1.30) !!!

16. Example problems will be
posted separately.
