HENDERSON-
HASSELBALCH
EQUATION
pH of solutions of weak acids cont’d
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Henderson-Hasselbalch Equation: Understanding pH in Weak Acids and Buffers, Lecture notes of Chemistry
The Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of a weak acid and its conjugate base. It includes examples of calculating pH and pKa/pKb for a weak acid, as well as an explanation of how buffers work and their preparation.
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HENDERSON-
HASSELBALCH
EQUATION
pH of solutions of weak acids cont’d
Example pH of solutions of weak acids
- The K a
for a weak acid, is 1.6 × 10
. The
molarity of acid is 10
M. What are the:
A) pH.
B) Calculate pK
a
and pK
b.
A) pH = ½ ( pK
a
+ p [HA])
pK
a
= - log K
a
pK
a
= - log K
a
= - log 1.6 × 10
pK
a
Henderson-Hasselbach equation
- A buffer is a solution that can resist changes in pH when
small amounts of acid or base is added.
- It is a mixture of a weak acid and its salt of a strong base
(an acidic buffer) OR it is a mixture of a weak base and it’s
salt of a strong acid (a basic buffer).
Handerson-Hasselbalch equation cont’ed
For acidic: HA H
+ A
K
a
[H
] = K
a
Log[H
] = Log K
a
+ Log
- Log[H
] = - Log K
a
- Log
pH = pK
a
+ Log
[H
] [A
- ]
[HA]
[A
- ]
[HA]
[A
- ]
[HA]
[A
- ]
[HA]
[A
- ]
[HA]
Handerson-Hasselbalch equation cont’ed
- When the condensation of conjugate acid = conjugate
base, pH = pK a
OR pOH = pK b
Buffers
- Buffer Capacity:
- It is the ability of buffer to resist changes in pH.
- It is the number of moles of H
ions that can be added to
one liter of the buffer that can decrease the pH by one unit
OR the number of moles of OH
- ions that can be added to
one liter of the buffer that can increase the pH by one
unit.
- Unit buffer capacity = mole.
Buffers cont’ed
When OH
are added it will react with the acetic acid:
CH 3 COOH + OH
- CH 3 COO - + H 2 O
Thus the buffer converted the free OH
- in the into water and salt
which does not affect the pH.
Preparation of buffers
- Example 1: What is the concentration of acetic acid and
acetate in 0.2 M acetate buffer, and which has a pH = 5
and pK a
Acetate buffer
Acetic acid + Acetate
HA + A
[HA] + [A
- ] = 0.2 M
Let us assume [A
- ] = y
Since [HA] + [A
- ] = 0.2 M
[HA] = 0.2 – y
Ka =
1.7 × 10
- = [(1 × 10 - )(y)] / (0.2 – y)
1.7 × 10
- (0.2 – y) = 1 × 10 - y
(3.4 × 10
- ) – (1.7 × 10 - y) = 1 × 10 - y
3.4 × 10
- = 1 × 10 - y + 1.7 × 10 - y
3.4 × 10
- = 2.7 × 10 - y
y = (3.4 × 10
- / 2.7 × 10 - )
y = 0.126 M = [A
- ]
[HA] = 0.2 – 0.126 = 0.074 M
[H
] [A
- ]
[HA]
Example 2
- Describe the preparation of 3 L of 0.2 M acetate buffer.
Starting from solid sodium acetate trihydrate (A
- ), Mwt =
136 and a 1 M solution of acetic acid (HA) the pK a
The concentration of [A
] = 0.126 M, [HA] = 0.074 M in
0.2 M solution in 1 L.
The no. of moles in buffer = 3 × 0.2 = 0.6 moles
SINCE A
- is solid the wt needed = M × Mwt = 0.378 × 136
= 51.4 g
The volume if HA needed = no. of moles / M = 0.222 / 1 =
0.222 L = 222 ml
51.4 g of solid sodium acetate trihydrate is added to 222
ml of acetic acid and the volume is brought up to 3 L.