Extraction is one the most common separation methods and it is based on the distribution equilibria — Organic Chemistry Chemistry Question
Separation by Extraction
Extraction is one the most common separation methods and it is based on the distribution equilibria of a substance between two immiscible liquids whose densities differ appreciably so that they are separated easily after mixing.
The more usual case is the extraction of an aqueous solution with an organic solvent, whereupon the inorganic ions and the polar organic compounds are found mainly in the aqueous phase and the non-polar organic compounds distribute in the organic phase. Inorganic ions may be reacted with an appropriate reagent to yield a non–polar compound that distributes in the organic phase.
When a species S (solute) is distributed between two solvents 1 and 2, then we have the following equilibrium:
(S)1 <=> (S)2
with the distribution coefficient KD given by:
KD = (aS)2 / (aS)1 (1)
where (aS)1 and (aS)2 are the activities of S in phases 1 and 2. For a given system of solvents and species S, KD depends practically solely on temperature. Separations by extraction are commonly performed with a separatory funnel (Fig. 1) [VISUAL], a common and easy to use glass apparatus found in any chemical laboratory.
Equation 1 is valid only if the solute S is present in both phases in the same form. Otherwise, if any dissociation, dimerization, complexation of the solute takes place, the distribution ratio, D, is used instead, which is given by:
D = (cS)2 / (cS)1 (2)
where (cS)1 and (cS)2 are the analytical concentrations of S in phases 1 and 2 (rather than equilibrium concentrations of given species).
By convention, when one of the two solvents is water, Equation 2 is written with the aqueous concentration in the denominator and the organic solvent concentration in the numerator. D is a conditional constant dependent on a variety of experimental parameters like the concentration of S and of any other species involved in any equilibria with S in either phase and most likely on the pH of the aqueous phase (e.g. if S participates in any acid–base type equilibrium).
If w0 (g) of S is initially present in a volume V1 (cm3) of solvent 1 and S is extracted successively with equal volume fractions V2 (cm3) of solvent 2, the quantity wn of S that remains in phase 1 after n such extractions is given by:
wn = w0 * (V1 / (D * V2 + V1))^n (3)
or
fn = wn / w0 = (V1 / (D * V2 + V1))^n (4)
where fn is the fraction of S that remains in solvent 1 after n extractions.
One can derive from equations (3) and (4) that for a given volume of extractant, successive extractions with smaller individual volumes of extractant are more efficient than with all the volume of the extractant.
Prove equation (3).
Model Answer
Starting with an amount w0 of S in phase 1, after the extraction this amount is distributed between the two phases as follows:
w0 = (cS)1 * V1 + (cS)2 * V2
Since D = (cS)2 / (cS)1, we have:
w0 = (cS)1 * V1 + D * (cS)1 * V2 = (D * V2 + V1) * (cS)1
Therefore, after removing phase 2, the remaining amount of S in phase 1 is:
w1 = (cS)1 * V1 = V1 * w0 / (D * V2 + V1)
By repeating extraction with a fresh portion of volume V2 of phase 2, the amount w1 of S is similarly distributed. After removing phase 2, the remaining amount of S in phase 1 is:
w2 = (cS)1 * V1 = V1 * w1 / (D * V2 + V1) = w0 * (V1 / (D * V2 + V1))^2
and so on. Therefore, after n extractions with a fresh portion of volume V2 of phase 2, the remaining amount of S in phase 1 will be:
wn = w0 * (V1 / (D * V2 + V1))^n
Substance S is distributed between chloroform and water with a distribution ratio D = 3.2. If 50 cm3 of an aqueous solution of S is extracted with
(a) one 100 cm3 portion, and
(b) four 25 cm3 portions of chloroform, calculate the percentage of S which is finally extracted in each case.
Model Answer
(a) The remaining fraction of S after 1 extraction with 100 cm3 of chloroform is calculated using Equation (4):
f1 = w1 / w0 = 50 / (3.2 * 100 + 50) = 0.135
Therefore, the percentage of S extracted is 100 - 13.5 = 86.5%
(b) The remaining fraction of S after 4 extractions with 25 cm3 of chloroform each time is similarly calculated:
f4 = w4 / w0 = (50 / (3.2 * 25 + 50))^4 = 0.022
Therefore, the percentage of S extracted is 100 - 2.2 = 97.8%.
The result is indicative of the fact that successive extractions with smaller individual volumes of extractant are more efficient than a single extraction with all the volume of the extractant.
What is the minimum number of extractions required for the removal of at least 99 % of substance X from 100 cm3 of an aqueous solution containing 0.500 g of X, if each extraction is carried out with 25.0 cm3 of hexane and the distribution coefficient is 9.5?
Model Answer
Using Equation (4), we have:
fn = wn / w0 = (V1 / (D * V2 + V1))^n
Since we want to remove at least 99% of substance X, the remaining fraction fn must be at most 1% (or 0.01):
0.01 = (100.0 / (9.5 * 25.0 + 100.0))^n
0.01 = 0.2963^n
Taking logarithms on both sides:
log(0.01) = n * log(0.2963)
-2.00 = n * (-0.5283)
n = 3.78
Therefore, at least 4 extractions are required.
The weak organic acid HA with dissociation constant Ka (in water) is distributed between an organic solvent and water. If the only extractable species is the undissociated species HA with a distribution coefficient KD and this is the only existing form of the acid in the organic phase, derive an expression showing the dependence of the distribution ratio D on [H+] of the aqueous phase and draw conclusions from the expression.
Model Answer
The equilibria involved are represented schematically as follows:
[VISUAL]
We have the following relations (where subscripts w and o denote concentrations in the aqueous and organic phases, respectively):
KD = [HA]o / [HA]w
Ka = [H+]w * [A-]w / [HA]w
D = (cHA)o / (cHA)w = [HA]o / ([HA]w + [A-]w)
Combining these equations, we obtain:
D = KD * [H+]w / ([H+]w + Ka) (1.5)
This equation predicts that if [H+]w >> Ka (strongly acidic aqueous phase), then D ≈ KD (i.e. D acquires its maximum possible value) and the acid is extracted (prefers to stay) in the organic phase. On the other hand, if [H+]w << Ka (strongly alkaline aqueous phase), we have D ≈ KD * [H+]w / Ka, and because of the small value of D, the acid is then retained (prefers to stay) in the aqueous phase. In this way, by regulating the pH of the aqueous phase, the course of extraction can be shifted towards the desired direction.
Benzoic acid and phenol are weak monoprotic organic acids with dissociation constants 6.6×10^-5 and 1×10^-10, respectively. Only the undissociated form of both compounds can be extracted from an aqueous solution with diethylether.
(a) Draw a plot showing how the ratio D/KD depends on pH for each compound,
(b) based on these plots propose a method for the separation of a mixture of these two compounds.
Model Answer
(a) By using the previously derived Equation 1.5, we obtain the following plots of the D/KD ratio vs. pH:
[VISUAL]
(b) From these plots it is clear that in the pH region of 7–8, the distribution ratio for benzoic acid will be practically 0 (as it is deprotonated into benzoate), whereas that of phenol (which remains protonated and neutral) will acquire its maximum possible value. Therefore, phenol can be efficiently separated from an aqueous solution of both compounds by extraction with diethylether, provided that the pH of the solution has been adjusted to the range of 7 to 8 (e.g., by adding an excess of NaHCO3).
8–Hydroxyquinoline, C9H6(OH)N (symbolized as: OxH), known also as “oxine”, forms in acidic solutions the cation C9H6(OH)NH+ (OxH2+) and in alkaline solutions the anion C9H6(O-)N (Ox-). If chloroform extracts only the neutral molecule of oxine with a distribution coefficient KD = 720,
(a) derive an expression showing the dependence of the distribution ratio, D, of 8–hydroxyquinoline on [H+] of the aqueous phase,
(b) draw a plot showing the dependence of D on pH of the aqueous phase,
(c) calculate the pH at which D is maximized.
The consecutive dissociation constants of the cationic acid C9H6(OH)NH+ are: K1 = 1×10^-5, and K2 = 2×10^-10 (see equilibrium reactions below).
[VISUAL]
Model Answer
(a) The equilibria involved are represented schematically as follows:
[VISUAL]
We have the expressions:
D = (cOx)o / (cOx)w = [OxH]o / ([OxH2+]w + [OxH]w + [Ox-]w)
KD = [OxH]o / [OxH]w = 720
K1 = [OxH]w * [H+]w / [OxH2+]w = 1×10^-5
K2 = [Ox-]w * [H+]w / [OxH]w = 2×10^-10
Combining all four equations, we obtain the sought-for expression:
D = KD / ([H+]w / K1 + 1 + K2 / [H+]w)
(b) Using the derived equation, we obtain the following D-pH plot:
[VISUAL]
(c) We calculate the 1st and 2nd derivative of the denominator, F([H+]w) = [H+]w / K1 + 1 + K2 / [H+]w.
The 1st derivative is:
F'([H+]w) = 1 / K1 - K2 / [H+]w^2
And the 2nd derivative is:
F"([H+]w) = 2 * K2 / [H+]w^3
Since F"([H+]w) > 0 for all positive [H+]w, F([H+]w) is minimized (and thus D is maximized) when F'([H+]w) = 0:
1 / K1 - K2 / [H+]w^2 = 0 ⇒ [H+]w^2 = K1 * K2 = (1×10^-5) * (2×10^-10) = 2×10^-15
[H+]w = sqrt(2×10^-15) = 4.47×10^-8 M
pH = -log(4.47×10^-8) = 7.35