Henderson-Hasselbalch Equation: Understanding pH in Weak Acids and Buffers, Lecture notes of Chemistry

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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.