Solubility: Combined effect of pH and solvents, distribution of solutes between immiscible solvents, extraction and MCQs for GPAT, NIPER, Pharmacist and Drug Inspector exam
Combined Effect of pH and Solvents:
Stockton and Johnson and Higuchi studied the effect of an increase of alcohol concentration on the dissociation constant of sulfathiazole, and Edmonson and Goyan investigated the effect of alcohol on the solubility of phenobarbital.
Schwartz determined the solubility of phenytoin as a function of pH and alcohol concentration in various buffer systems and calculated the apparent dissociation constant. Kramer and Flynn examined the solubility of hydrochloride salts of organic bases as a function of pH, temperature, and solvent composition. They described the determination of the pKa of the salt from the solubility profile at various temperatures and in several solvent systems. Chowhan measured and calculated the solubility of the organic carboxylic acid naproxen and its sodium, potassium, calcium, and magnesium salts.
Distribution of Solutes between Immiscible Solvents:
If an excess of liquid or solid is added to a mixture of two immiscible liquids, it will distribute itself between the two phases so that each becomes saturated. If the substance is added to the immiscible solvents in an amount insufficient to saturate the solutions, it will still become distributed between the two layers in a definite concentration ratio.
If C1 and C2 are the equilibrium concentrations of the substance in Solvent1 and Solvent2, respectively, the equilibrium expression becomes
C1/C2 = K (1)
The equilibrium constant, K, is known as the distribution ratio, distribution coefficient, or partition coefficient. Equation (1), which is known as the distribution law, is strictly applicable only in dilute solutions where activity coefficients can be neglected.
Knowledge of partition is important to the pharmacist because the principle is involved in several areas of current pharmaceutical interest. These include preservation of oil–water systems, drug action at nonspecific sites, and the absorption and distribution of drugs throughout the body.
Extraction:
To determine the efficiency with which one solvent can extract a compound from a second solvent—an operation commonly employed in analytic chemistry and in organic chemistry—we follow Glasstone. 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)/V2 g/mL. The distribution coefficient is thus
K = (w1/v1)/(w- w1)v2 (2)
Or
w1 = w.K v1/ K v1+ v2 (3)
The process can be repeated, and after n extractions,
Wn = w(K v1/ K v1+ v2) (4)
By 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. The development just described assumes complete immiscibility of the two liquids. When ether is used to extract organic compounds from water, this is not true; however, the equations provide approximate values that are satisfactory for practical purposes. The presence of other solutes, such as salts, can also affect the results by complexing with the solute or by salting out one of the phases.
Multiple choice questions:
1.Who studied the effect of an increase of alcohol concentration on the dissociation constant of sulfathiazole?
a)Stockton
b)Johnson
c)Higuchi
d)all of these
2.Who investigated the effect of alcohol on the solubility of phenobarbital?
a)Edmonson
b)Goyan
c)Higuchi
d)a and b
3.Schwartz determined the solubility of phenytoin as a function of pH and alcohol concentration in various buffer systems and calculated the apparent
a)viscosity
b)dissociation constant
c)solubility
d)all of these
4.Kramer and Flynn examined the solubility of hydrochloride salts of organic bases as a function of
a)pH
b)temperature
c)solvent composition
d)all of these
5.Chowhan measured and calculated the solubility of the organic carboxylic acid naproxen and its sodium, potassium, calcium, and magnesium salts.
a)true
b)false
6.If C1 and C2 are the equilibrium concentrations of the substance in Solvent1 and Solvent2, respectively, the equilibrium expression becomes
a)C1/C2 = K
b)C2/C1 = K
c)C1/C = K
d)C/C2 = K
7.The equilibrium constant, K, is known as
a)distribution ratio
b)distribution coefficient
c)partition coefficient
d)all of these
8.C1/C2 = K This equation is known as
a)distribution ratio
b)distribution law
c)distribution coefficient
d)partition coefficient
9.Wn = w(K v1/ K v1+ v2) By use of this equation, it can be shown that most efficient extraction results when n is large and V2 is small.
a)true
b)false
10.Knowledge of partition is important to the pharmacist because the principle is involved in several areas of current pharmaceutical interest. These include
a)oil–water systems
b)drug action at nonspecific sites
c)the absorption and distribution of drugs throughout the body
d)all of these
11.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)/V2 g/mL. The distribution coefficient is thus
a)K = (w1/v1)/(w- w1)v2
b)w1 = w.K v1/ K v1+ v2
c)both of these
d)none of these
12.Distribution law, is strictly applicable only in ____ solutions where activity coefficients can be neglected.
a)dilute
b)concentrated
c)heterogenous
d)homogenous
13.If an excess of liquid or solid is added to a mixture of two immiscible liquids, it will distribute itself between the two phases so that each becomes saturated.
a)true
b)false
14.If the substance is added to the immiscible solvents in an amount insufficient to saturate the solutions, it will still become distributed between the two layers in a indefinite concentration ratio.
a)true
b)false
15.Distribution law is given as
a)C1/C2 = K
b)K = (w1/v1)/(w- w1)v2
c)w1 = w.K v1/ K v1+ v2
d)C2/C1 = K
Solution:
- d)all of these
- d)a and b
- b)dissociation constant
- d)all of these
- a)true
- a)C1/C2 = K
- d)all of these
- b)distribution law
- a)true
- d)all of these
- c)both of these
- a)dilute
- a)true
- b)false
- a)C1/C2 = K
Reference:
- Martins Physical Pharmacy, 6th edition 2011, page no. 344-354.
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