Research: Science and Education
Effect of Mutual Solubility of Solvents in Multiple Liquid–Liquid Extraction Tadeusz Michalowski Department of Analytical Chemistry, Institute of Inorganic Technology, Technical University of Crakow, ul. Warszawska 24, 31-155 Kraków, Poland;
[email protected] Liquid–liquid extraction in systems consisting of two solvents is usually described using an oversimplified model based on the following assumptions (1–5): 1. The solvents are completely immiscible. 2. The volumes of the phases obtained after mixing the solvents (or the related solutions) are additive. 3. The composition of both phases is constant, except for the species transferred.
In particular, the last assumption implies that the pH value of the aqueous phase should be constant. However, it is evident that the extraction of a component exhibiting acid–base properties must affect the pH and composition (speciation) of the system, particularly if there is no buffering action. This has been found for some complex redox systems (6 ), where disproportionation (dismutation) reactions occur (6, 7 ). The problem raised here is not discussed in physical chemistry textbooks when the mutual solubility of liquids is considered. The question also has some relevance to analytical chemistry where multiple liquid–liquid extraction is employed for preconcentration purposes and is involved in some spectrophotometric determinations. This paper refers principally to the process of multiple extraction. Examination of literature data (4 ) shows that the mutual solubilities of water and certain organic solvents can be high, as for example for the isoamyl alcohol-plus-water system. The high solubilities are accompanied by definite
changes in the physicochemical properties of the phases formed—for example, in dielectric permeability and in the solubilities of any substances added to the system. The mutual solubilities are affected to an appreciable extent by the presence of the substances dissolved in the phases thus formed. This effect is particularly important for systems of concentrated (aqueous) solutions of electrolytes and is enhanced during multiple extractions. Consider the solubility diagram (Fig. 1) for a two-phase system composed of two partially miscible substances (solvents), A and B; we then have phase 1 (A + B), composed of A saturated with respect to B and phase 2 (B + A), composed of B saturated with respect to A.
Let mA g of A be mixed with mB g of B at a temperature T = T * (Fig. 1) Denoting the percentage contents of A in the mixture by a = 100 × mA/(mA + mB), we get mB/mA = (100 – a)/a. The resulting phase 1 (mass m1) contains p% (m/m) of A and phase 2 (mass m2) contains q% (m/m) of A. From the equations of mass balance for A and B pm1 + qm2 = 100mA; (100 – p)m1 + (100 – q)m2 = 100mB (1) we obtain the relations and
m1 = [(100 – q)mA – qmB]/( p – q) m2 = [ pmB – (100 – p)mA]/( p – q)
and then the formula m1/m2 = (a – q)/(p – a)
(2a)
expressing the well-known “lever-arm” principle m1(p – a) = m2(a – q)
Temperature
Tk C
(2b)
is obtained. Let us now consider a system composed of water (A) and an organic solvent (B), partially miscible with water. After mA g of A and mB g of B are shaken together, equilibrium is established, with
T*
m11 g of the aqueous phase (phase 1) containing p1% of B and 100% A p 0% B
a
q
0% A 100% B
Mole Fraction (%) Figure 1. The solubility diagram related to the system formed of two partially miscible liquid substances (A and B) under isobaric conditions. The points above isobaric line (c) express the monophase system; below c, a two-phase system (A + B and B + A) is formed. At temperatures T > Tk, the monophase system is formed, independent of the mixture composition.
m21 g of the organic phase (phase 2), containing p2% of A Then, after the mixture is allowed to stand, the organic phase is separated and the remaining aqueous phase is shaken with another portion mB g of solvent B. Suppose the procedure is repeated i times. If equilibrium is attained, the masses of the aqueous (m1i) and organic (m2i) phases after the ith addition of the organic solvent B to water or the aqueous phase (phase 2) can be calculated. It should be noted that masses (not volumes) are treated as strictly additive.
JChemEd.chem.wisc.edu • Vol. 79 No. 10 October 2002 • Journal of Chemical Education
1267
Research: Science and Education
After the addition of the first portion of B one gets (Fig. 2) and
(100 – p1)m11 + p2m21 = 100mA
mA-mA
p1m11 + (100 – p2)m21 = 100mB
mA,
(compare with eq 1) and then:
,
mB
m B, m B-m B,
2 (B + trace A)
m11 = [100mA – p2(mA + mB)]/(100 – p1 – p2)
(3)
m21 = [100mB – p1(mA + mB)]/(100 – p1 – p2)
(4)
A (water)
phases 1 and 2
B (organic)
Figure 2. Distribution of water (A) and organic solvent (B) within phases 1 and 2 (1st extraction). Phase 1, with mass m11, contains mB′ = (p1/100)m11 g of B and (100 – p1)/100m11 g of A; phase 2, with mass m21, contains mA′ = (p2/100)m21 g of A and (100 – p2)/100m21 g of B.
After the addition of ith (i ≥ 2) portion of B, we have and
1 (A + trace B)
mA
(100 – p1)m1i + p2m2i = (100 – p1)m A,i-1 p1m1i + (100 – p2)m2i = 100m B + p1m A,i-1
and then m1i = m 1,i-1 – mBp2/(100 – p1 – p2)
(5)
m2i = mB(100 – p1)/(100 – p1 – p2)
(6)
From eqs 5 and 6 it results that the masses of the aqueous phase in successive stages of the extraction decrease and form a sequence (arithmetical progression) of numbers, whereas the masses of the organic phase remain constant for i ≥ 2. From eq 5 we get m1i = m1,i-j – jmBp2/(100 – p1 – p2)
(7)
Setting i – j = 1, we have j = i – 1; that is, m1i = m11 – (i – 1)mBp2/(100 – p1 – p2)
(8)
The aqueous phase finally disappears (m1k = 0) at i = k ≥ r (more exactly, for k ≥ mod(r + 1)), where r = (mA/mB)(100 – p2)/p2
(9)
The relationship in eq 9 is obtained from eqs 8 and 3. For example, 5.67 g of H2O dissolves in 100 mL of cyclohexanol (C6H11OH, d = 0.945 g/cm3 at 20 °C) (4 ); that is, then
p2 = 5.67/(100 × 0.945 + 5.67) = 5.66% r = (mA/mB)(100 – 5.66)/5.66 = 16.67
1268
at mB = mA, mod(r) = 17. For a CCl4 + H2O system we get (4 ) then
p2 = 0.08/(100 × 1.595 + 0.08) = 0.050% r = (mA/mB)(100 – 0.050)/0.050 ≈ 2000
at mB = mA. The calculations and comments presented above show the necessity of using mutually saturated phases (1 and 2) or solutions prepared on the basis of the phases. This makes it possible to avoid the effect of a decrease in the volume of the aqueous phase in the course of the successive extractions (3). Literature Cited 1. Kertes, A. S.; Marcus, Y. Solvent Extraction Research; Wiley: New York, 1969. 2. De, A. K.; Khopkar, S. M.; Chalmers, R. A. Solvent Extraction of Metals; Van Nostrand Reinhold: London, 1970. 3. Zolotov, Yu. A.; Kuzmin, N. M. Extraction Concentration; Khimia: Moscow, 1971; in Russian. 4. Korenman, I. M. Extraction in Analysis of Organic Substances; Khimia: Moscow, 1977; in Russian. 5. Korenman, I. M. Zh. Anal. Khim. 1982, 37, 301. 6. Michalowski, T.; Lesiak, A. J. Chem. Educ. 1994, 71, 632–636. 7. Michalowski, T. J. Chem. Educ. 1994, 71, 560–562.
Journal of Chemical Education • Vol. 79 No. 10 October 2002 • JChemEd.chem.wisc.edu
