Buffer & Isotonic Solutions, Lecture notes of Physical Chemistry

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BUFFERS

DEFINITION

Buffers are compounds or mixtures of

compounds that by their presence in the

solution resist changes in the pH upon the

addition of small quantities of acid or alkali.

TYPES OF BUFFERS

Generally buffers are of two types;

• Acidic buffers
• Basic buffers

ACIDIC BUFFERS An acidic buffer is a combination of weak acid and its salt with a strong base. ➢ Weak acid & salt with strong base (conjugate base). Examples:

• CH 3 COOH / CH 3 COONa
• H 2 CO 3 / NaHCO 3
• H 3 PO 4 / NaH 2 PO 4
• HCOOH / HCOONa

PHOSPHATE BUFFERS (DOUBLE SALT BUFFERS)

Besides the two general types of buffers (i.e. Acidic & basic), a third appears to exist. This is buffer system composed of two salts:

• Monobasic potassium phosphate (KH 2 PO 4 )
• Dibasic potassium phosphate (K 2 HPO 4 ).

BUFFER ACTION

• The resistance of a buffer solution to a change in pH is known as buffer action. Mechanism of action of acidic buffers:
• Consider a buffer system of CH 3 COOH (weak electrolyte) and CH 3 COONa (strong electrolyte). There will be a large concentration of Na + ions & CH 3 COO - ions and undissociated CH 3 COOH molecules. When an acid is added
• If a strong acid (HCl) is added in CH 3 COOH/CH 3 COONa buffer, the changes that will occur may be represented as: 3 3 3

_ _

MECHANISM OF ACTION OF ACIDIC BUFFERS When a base is added:

• If a strong base (NaOH) is added in CH 3 COOH/CH 3 COONa buffer, the changes that will occur may be represented as: 3 3 The hydroxyl ions yielded by the NaOH are therefore removed as water. The supply of hydrogen ions needed for this purpose being constantly provided by the dissociation of acetic acid. _ + _ + 2

MECHANISM OF ACTION OF BASIC BUFFERS

• Consider a buffer system of NH 4 OH (weak electrolyte) and NH 4 Cl (strong electrolyte). There will be a large concentration of NH 4 + ions, Cl - ions, and undissociated NH 4 OH molecules. When an acid is added
• If a strong acid (HCl) is added in NH 4 OH/NH 4 Cl buffer, the changes that will occur may be represented as: (^4 ) The hydrogen ions yielded by the HCl are therefore removed as water. The supply of OH ions needed for this is constantly provided by the ammonium hydroxide.
  • _ + _ 2

MECHANISM OF ACTION OF PHOSPHATE BUFFERS

In KH

2

PO

4

/K

2

HPO

4

buffer system, H

2

PO

4

serves as weak acid and

HPO

4

• 2

serves as conjugate base. When hydronium ions are added,

then-

HPO

4

• 2

+ H

3

O

→ H

2

PO

4

+ H

2

O

When hydroxyl ions are added to this buffer, the following reaction

takes place;

H

2

PO

4

+ OH

→ HPO

4

• 2

+ H

2

O

BUFFER EQUATION (Henderson – Hasselbalch equation)

A weak acid is only slightly dissociated and its dissociation is further depressed by the addition of the salt (XA) which provides A

• ion (common ion effect) as a result the equilibrium concentration of the unionized acid is nearly equal to the initial concentration of the acid. The equilibrium concentration of A

is assumed to be equal to the initial concentration of the salt added since it is completely dissociated. Therefore, in above equation ( 1 ), we represent concentration of A- by salt concentration.

[H

] = K a [Acid] / [Salt] --------- (2) Taking log on both sides, we get: log [H

] = log K a

• log [Acid] / [Salt] Multiplying both sides by (–ve) sign,

• log [H
  • ] = - log K a - log [Acid] / [Salt] As - log[H
  • ] = pH & - logK a = pK a pH = pK a
    • log [Acid] / [Salt] **pH = pk a

• log [Salt] / [Acid] ---------- (3)**

Eq. ( 3 ) is called as Henderson – Hasselbalch equation. It helps in

calculating the pH value of buffer solution, if the concentrations of

acid as well as that of the salt are known.

FOR BASIC BUFFERS

A weak base is only slightly dissociated, and its dissociation is

further depressed by the addition of the salt (BA) which

provides B

ion (common ion effect) as a result the

equilibrium concentration of the unionized base is nearly

equal to the initial concentration of the base. The equilibrium

concentration of B

is assumed to be equal to the initial

concentration of the salt added since it is completely

dissociated. Therefore, in above equation ( 1 ), we represent

concentration of B

by salt concentration.

[OH

• ] = K b [Base] / [Salt] --------- (2) Taking log on both sides, we get: log[OH

] = log K b

• log [Base] / [Salt] multiplying both sides by (–ve) sign,

• log[OH

] = - log K b

• log [Base] / [Salt] As - log[OH

] = pOH & - logK b = pK b pOH = pK b

• log [Base] / [Salt] or

pOH = pK

b

+ log [Salt] / [Base] ---------- (3)