PHYSICAL CHEMISTRY IN BRIEF

CHAP. 7: PHASE EQUILIBRIA [CONTENTS] 215

can write the following formula for the residual amountnr

nr=n 0 −np=n 0

(

V 1

V 1 +Kc, 3 V 2

)k

. (7.61)

Note:Nernst’s distribution law (7.58) assumes a constant value of the distribution

coefficient. This assumption is justified if the solubility of component 3 in both immiscible

liquids is low. At infinite dilution of the component in both liquid phases, the distribution

coefficient acquires the limiting value which is given by the ratio of the limiting activity

coefficients [compare with (7.59)].

Example

1000 cm^3 of water containsn 0 = 0. 0001 moles of iodine. What amount of iodine remains in

the aqueous solution after extraction by carbon disulphide if we use: a) 50 cm^3 in a one-off

extraction, b) 10 cm^3 in an extraction repeated five-times. Nernst’s distribution coefficient is

Kc,I 2 =cCSI 22 /cHI 22 O= 600.

Solution

a) from (7.61) we obtain

nr= 0. 0001

1000

1000 + 600× 50

= 3. 33 × 10 −^6.

b)

nr= 0. 0001

(

1000

1000 + 600× 10

) 5

= 5. 94 × 10 −^9.

In case b), i.e. after repeating the extraction five-times, about 500-times less iodine remains in

water than in the case of the one-off extraction.