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.