Solubility part 9

The meaning of C in equation (9-18) is understood readily by considering a simple illustration. Suppose one begins with 1 liter of oil and 1 liter of water, and after benzoic acid has been distributed between the two phases, the concentration   of benzoic acid in the oil is 0.01 mole/liter and the concentration Cw of benzoic acid in the aqueous phase is 0.01 mole/liter. Accordingly, there is 0.02 mole/2 liter or 0.01 mole of benzoic acid per liter of total mixture after distribution equilibrium has been attained. Equation (9-18) gives:
  The concentration, C, obviously is not the total concentration of the acid in the mixture at equilibrium but, rather, twice this value. C is therefore seen to be the concentration of benzoic acid in the water phase (or the oil phase) before the distribution is carried out.
Second, let us now consider the case in which the solute is associated in the organic phase and exists as simple molecules in the aqueous phase. If benzoic acid is distributed between benzene and acidified water, it exists mainly as associ¬ated molecules in the benzene layer and as undissociated molecules in the aqueous layer.
The equilibrium between simple molecule HA and associated molecules (HA)n in
Where (HA)n are the associated molecules and n(HA) are the simple molecules, and the modified distribution constant becomes:
Extraction
To determine the efficiency with which one solvent can extract a compound from a second solvent which is an operation commonly employed in analytical chemistry and in organic chemistry. Suppose that w grams of a solute is extracted repeatedly from V1 mL of one solvent with successive portions of V2 mL of a second solvent, which is immiscible with the first. Let w1 be the weight of the solute remaining in the original solvent after extracting with the first portion of the other solvent. Then, the concentration of solute remaining in the first solvent is (w1/V1)g/mL, and the concentration of the solute in the extracting solvent is (w–w1)/V2g/mL. the distribution coefficient is thus:    or
The process can be repeated, and after n extractions
By the use of this equation, it can be shown that most efficient extraction results when n is large and V2 is small, in other words, when a large number of extractions are carried out with small portions of extracting liquid.
Example: The distribution coefficient for iodine between water and carbon tetrachloride at 25°C is  How many grams of iodine are exteracted from a solution in water containing 0.1 g in 50 mL by one extraction with 10 mL of CCl4? How many grams are extracted by two 5-mL portions of CCl4?
Thus, 0.0011 g of iodine remains in the water phase, and the two portions of CCl4 have extracted: