Solubility and Distribution Phenomena PDF

# CHAPTER 11 Solubility and Distribution Phenomena Solubility is a phenomenon which involves formation of a uniform homogenous dispersion upon interaction of two or more substances. It is dependent on many factors like temperature, nature of solute and solvent under study, pH, pressure etc. Quanti...

# CHAPTER 11 Solubility and Distribution Phenomena Solubility is a phenomenon which involves formation of a uniform homogenous dispersion upon interaction of two or more substances. It is dependent on many factors like temperature, nature of solute and solvent under study, pH, pressure etc. Quantitatively the solubility is expressed in terms of concentration or amount of solute to the volume of solution. Based on states of matter, a total of nine possible types mixtures can be formed as shown in Table 11.1. | Nature of solute | Nature of solvent | Example | | ---------------- | ------------------- | ------------------- | | Gas | Gas | Air | | Gas | Liquid | Oxygen in water | | Gas | Solid | Hydrogen in palladium | | Liquid | Liquid | Ethanol in water | | Liquid | Solid | Mineral oil in paraffin | | Liquid | Gas | Moisture in air | | Solid | Liquid | Sugar solution | | Solid | Solid | Various alloys | | Solid | Gas | Iodine vapour in air | ## Solubility and Distribution Phenomena ### (a) Polar Solvents: The polarity of solvent depends up on its dipole moment. For example water has its dielectric constant value 80 and hence it falls under the category of strongly polar solvents. The solubility of compounds in solvents can be generalized under statement "like dissolves in like", i.e., polar solvents have strong solubilizing power for polar substances or ionic solutes. Ionic solutes include salts like sodium chloride, potassium chloride etc. The solubility of a compound also depends on ability to form hydrogen bonds. Compounds having high dipole moment tend to form hydrogen bonds to a greater extent and hence show higher solubility in polar solvents. * **Alcohol:** ``` R-O-----H- О-----Н - О | H ``` * **Aldehyale:** ``` R-C=O-----H-O-----O = C - R | H ``` * **Ketone:** ``` R > C = O-----H-O----- O = C R ``` Water dissolves phenols, alcohols, aldehydes ketones that can form hydrogen bonds with water molecules. When polar solvents like alcohol (lower alcohols) are added to water the hydrogen bonds between water molecules are replaced partly by the hydrogen bonds between alcohol and water molecules. Apart from above mentioned factors the other parameters which play a key role in deciding the solubility include, ratio of polar to non polar groups of the molecule. The solubility is inversely proportional to the carbon chain length. E.g., The solubility of alcohols decreases with increasing carbon chain length. Branching of carbon chain increases the solubility of compound compared to its straight chain compound. ### (b) Non-polar Solvents: Non-polar solvents have the capacity of dissolving non polar compounds. Generally oils, fats come under this category. They possess a very low dielectric constat value generally ranging from 0 - 10. ## Effect of Temperature With the increase in temperature solubility of most of gases decreases owing to the greater tendency of gases to expand. For the same reason gas perfumes, body sprays etc are exercised with caution in warm and hot climates. These containers should not be disposed in fire as the gas expands rapidly and results in bursting of container. ## Salting Out Introduction of electrolyte to solution that has dissolved gas results in liberation of gases from the solution. This phenomenon is called "Salting Out". This is due to the higher affinity of electrolyte to water molecules, which reduces the density of aqueous environment adjacent to the gas molecules. ## Effect of Chemical Reaction Henry's law is strictly applicable to gas which do not react with the solvent. The reaction between the gas and solvent alters the solubility. E.g., Hydrogen chloride is about 10,000 times more soluble in water than is oxygen. ## Solubility of Liquids in Liquids Quite often we come across the solubility of liquids in liquids in pharmaceutical solutions. These include preparation of elixirs, aromatic waters, Medicated oils etc. As discussed earlier solutions can be classified into : * **Ideal Solutions:** In which binary solutions obey Raoult's law over the whole range of compositions. * **Real Solutions:** These are the binary solutions which show either positive or negative deviation from Raoult's law. If one compound shows positive deviation then other component also tends to show positive deviation. Positive deviations leads to decrease in solubility due to formation of dimers or polymers. Broadly one can classify liquid-liquid systems into three categories: * Complete miscibility * Partial miscibility * Immiscibility ## Solubility of Solids in Liquids This is yet another broader area of pharmaceutical interest. Broadly we can classify the solutions in to ideal solutions and real solutions. For ideal solutions the behaviour can be predicted in a satisfactory manner. ### Ideal Solutions: Various factors which affect the solubility of a solid include temperature, melting point, and molar heat of fusion AH, (heat absorbed when solid melts). For ideal solutions heat of fusion is equal to heat of solution, which is constant independent of the temperature. The equation derived from the thermodynamics for ideal solution is ``` log X ΔΗ(Τ, - T) -------- 2.303R TT, + constant where Κς, Τ₁ = saturation solubility at absolute temperature, T₁ Ks,T2 = saturation solubility at absolute temperature T2. ΔΗ = heat of solution. ``` ## Effect of Temperature As evident from the above equation the solubility of solid in a liquid depends on the temperature. If heat is absorbed in process of solution then AH will be positive and solubility of solute increases with increase in temperature. If a solute gives off heat during the process of solution, AH is negative and solubility decreases with increase in temperature. Most of the salts show positive AH values. Salts like calcium sulfate and calcium hydroxide show negative AH values. When heat is neither absorbed nor given off, the solubility is not affected by variation of temperature as is nearly the case with sodium chloride. ## Solubility of Strong Electrolytes Whenever a strong electrolyte is dissolved in a solution, we can observe the following: * Rise in the temperature of system, or exothermic process. * Decrease in the temperature of system or endothermic process. * No change in temperature of system. According to Lechaterlier principle, any system tends to adjust itself in a manner so as to counteract, any stress. So any change in the temperature of system will proceed in the direction, so as to nullify the change. ## Solubility of Slightly Soluble Electrolyte The solubility of slightly soluble electrolytes is described by solubility product, Ksp, of the compound. For example, consider silver chloride which when dissolved in excess exists in equilibrium as ``` Ag Cl solid Ag+ + Cl- ``` The equilibrium expression may be written as ``` K = [Ag+][cr] [AgCl solid] ``` As the concentration of [Ag ClSolid] is constant. Ksp = [Ag+] [Cl] Addition of sodium chloride to above silver chloride solution results in common ion effect. ``` [Ag+] [Cl] > Ksp ``` Some of the AgCl precipitates from the solution until equilibrium [Ag+] [Cl] = Ksp is re-established. ``` HP Solid HP Sol HP + H2O = H3O+ + P Sol ``` Equilibrium constant for Eq. (11.3) is ``` K= a [P] = [H3O+][-] [HP] Sol Ka [HP] Sol [H3O+ .....(11.2) .....(11.3) ``` ## Distribution Phenomenon When we add a solid or liquid to a system of two immiscible liquids, it will distribute itself between the two phase in a definite concentration ratio. The two immiscible liquids may be oil and water system or organic and aqueous phase system. Let the concentration of solute in organic phase be C Corg and aqueous phase be Caq at equilibrium. The distribution co-efficient K is the ratio of concentration of solute in organic phase to concentration of solute in aqueous phase. ``` Ko/w = Corg Caq or Kw/o = Caq Corg ``` K is also termed as partition coefficient, or distribution ratio. ## Applications of Distribution Coefficient 1. Absorption and distribution of drugs in the body. 2. Drug action at non-specific sites. 3. Preservation of oil-water system etc. ## Effect of Ionic Dissociation and Molecular Association on K: Some solutes may exist partly or wholly as associated molecules in one phase and may exist as dissociated ions in another phase. This phenomena leads to certain deviations from the original distribution law. If we consider the distribution of Benzoic acid in an oil/water system, we can clearly understand the above said deviation as shown in Fig. 11.5. ``` C=0 [HA]o H H Ka C OH Ka OH [HA]w + H+ K [HA]w [A-]w ``` ## Solubility and Distribution Phenomena So if we consider distribution coefficient, ``` K = [HA] Co [HA] Cw W ``` But Cw = [HA]+[A]w W (As certain molecules exist in dissociated form) So apparent distribution coefficient ``` [HA]o K' = -------- [HA]+[A]w ``` The dissociation constant of benzoic acid is given by ``` K= [H3O+][A] [HA] W ``` Eq. (11.4) and Eq. (11.5) can be combined to give ``` Ka+H3O+ Cw = K C a + K+1 C 173 .....(11.4) .....(11.5) [H3O+] ``` A plot of Ka+[H3O+] Cw line with slope equal to on y-axis and [H3O+] on x-axis yields a straight K+1 C and intercept = Ka C The oil/water distribution co-efficient is an indication of lipophilic or hydrophobic charactor of drug molecule. For the purpose of study of drug, passage through lipodial membrane, good correlation is observed by using octanol / water system. ## Solubility and Distribution Phenomena 3. What is the solubility of given substance at 20 °C in an ideal solution? The melting point of given substance is 80 °C and the molar heat of fusion is 4500 cal/mole. Solution: Data given Melting point = 80 °C = 353 °K Molar heat of fusion = 4500 cal/mole Equation for ideal solubility ``` log X ΔΗ Το - Τ -------- 2.303R TT ``` ``` 4500 り (353-293) log X 2.303×1.987 293×353 - X = 0.27. 2 ``` 4. A solid material is to be comminuted so as to increase its solubility by 10% i.e., S/S is 1.10. What is final radius of particles in cm, assuming that the surface tension of solid is 110 dynes/cm and volume per mole is 70 cm³. Temperature is 27 °C. Solution: Data given y = surface tension = 110 dynes/cm T = 300 K Volume per mole = 70 cm³ Equation ``` log s = S/ 2yV r= So 2.303 RTr ``` ``` 2×110×70 2.303×8.314×107×300×0.0414 = 6.4 x 10-6 cm = 0.064 µm. ```