Organic Chemistry

ether

water

52 CHAPTER 1 Electronic Structure and Bonding • Acids and Bases

1%

50%

90%

99%

3.2 4.2 5.2

pH = pKa − 2

pH = pKa − 1 pH = pKa + 1

pH = pKa pH = pKa^ +^2

pH

6.2 7.2

acidic form 10%

basic form

Figure 1.20N
The relative amounts of a
compound with a of 5.2 in the
acidic and basic forms at different
pH values.

pKa

The Henderson–Hasselbalch equation can be very useful in the laboratory when

compounds need to be separated from each other. Water and diethyl ether are not

miscible liquids and, therefore, will form two layers when combined. The ether layer

will lie above the more dense water layer. Charged compounds are more soluble in

water, whereas neutral compounds are more soluble in diethyl ether. Two compounds,

such as a carboxylic acid (RCOOH) with a of 5.0 and a protonated amine

with a of 10.0, dissolved in a mixture of water and diethyl ether, can be

separated by adjusting the pH of the water layer. For example, if the pH of the water

layer is 2, the carboxylic acid and the amine will both be in their acidic forms because

the pH of the water is less than the of both compounds. The acidic form of a car-

boxylic acid is neutral, whereas the acidic form of an amine is charged. Therefore, the

carboxylic acid will be more soluble in the ether layer, whereas the protonated amine

will be more soluble in the water layer.

For the most effective separation, it is best if the pH of the water layer is at least two

units away from the values of the compounds being separated. Then the relative

amounts of the compounds in their acidic and basic forms will be at least 100:1

(Figure 1.20).

pKa

RCOOH RCOO− + H+

acidic form basic form

RNH 3 RNH 2 + H+

+

pKa’s

(RNH 3 +) pKa

pKa

DERIVATION OF THE

HENDERSON–HASSELBALCH

EQUATION

The Henderson–Hasselbalch equation can be derived from the

expression that defines the acid dissociation constant:

Taking the logarithms of both sides of the equation and then, in

the next step, multiplying both sides of the equation by we

obtain

• 1,

Ka=

[H 3 O+][A-]

[HA]

and

Substituting and remembering that when a fraction is inverted,

the sign of its log changes, we get

pKa=pH+log

[HA]

[A-]

• log Ka=-log[H 3 O+]-log

[A-]

[HA]

log Ka=log[H 3 O+]+log

[A-]

[HA]