Chem 2 - Acid-Base Equilibria X: Buffers and the Henderson-Hasselbalch Equation

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.
