Physical Pharmacy, Solubility and Distribution.pdf

Physical Pharmacy Solubility & Distribution Definition & Importance Chapter 10| Solubility and Distribution Solubility and Distribution Phenomena : Solubility is defined in quantitative terms as; The concentration of solute in a saturated solution at a certain temperature, And in a qualitative way,...

Physical Pharmacy Solubility & Distribution Definition & Importance Chapter 10| Solubility and Distribution Solubility and Distribution Phenomena : Solubility is defined in quantitative terms as; The concentration of solute in a saturated solution at a certain temperature, And in a qualitative way, it can be defined as; The spontaneous interaction of two or more substances to form a homogeneous molecular dispersion Chapter 10| Solubility and Distribution Solutions and Solubility 1. A saturated solution is one in which the solute in solution is in equilibrium with the solid phase. 2. An unsaturated or subsaturated solution is one containing the dissolved solute in a concentration below that necessary for complete saturation at a definite temperature. 3. A supersaturated solution is one that contains more of the dissolved solute than it would normally contain at a definite temperature, were the undissolved solute present. Chapter 10| Solubility and Distribution Chapter 10| Solubility and Distribution Example: If we now slowly cool the mixture back to 25 °C, 9 g of glucose should precipitate from solution. Sometimes this happens immediately, but sometimes it takes a while for the glucose molecules to find their positions in a solid structure. In the time between the cooling of the solution and the formation of glucose crystals, the system has a higher amount of dissolved glucose (100 grams) than is predicted by the solubility limit at 25 °C (91 grams). Chapter 10| Solubility and Distribution Because the solution contains more dissolved solute than is predicted by the solubility limit, we say the solution is supersaturated. Chapter 10| Solubility and Distribution Chapter 10| Solubility and Distribution Some salts e.g. (sod thiosulfate) can be dissolved in large amounts at an elevated temperature and, upon cooling fail to crystallize from the solution (supersaturated). Factors affecting solubility: 1. physical and chemical properties of the solute and the solvent 2. Temperature of the solution 3. Pressure of the solution 4. pH of the solution 5. State of subdivision of the solute Chapter 10| Solubility and Distribution Solubility Expressions The solubility of a drug may be expressed in a number of ways. A. The solubility of a drug can be expressed in terms of: Molarity Molality Normality Mole fraction percentage (% w/w, % w/v, % v/v) B. The United States Pharmacopeia (USP) USP lists the solubility of drugs as the number of ml of solvent in which 1 g of solute will dissolve. E.g. 1g of boric acid dissolves in 18 mL of water, and in 4 mL of glycerin. C. The United States Pharmacopeia (USP) USP uses general description of substances solubility by the following terms: Chapter 10| Solubility and Distribution Solubility Definition Parts of Solvent Required for One Part of Solute Very soluble (VS) 10,000 Chapter 10| Solubility and Distribution Solute-Solvent interactions : Solute molecules are held together by certain intermolecular forces (dipoledipole, induced dipole-induced dipole, ion-ion, etc.), as are molecules of solvent. In order for dissolution to occur, these cohesive forces of like molecules must be broken and adhesive forces between solute and solvent must be formed. The solubility of a drug in a given solvent is largely a function of the polarity of the solvent. Chapter 10| Solubility and Distribution Solvent Water Glycols Methyl and ethyl alcohols Aldehydes, ketones, and higher alcohols, ethers, esters, and oxides Hexane, benzene, carbon tetrachloride, ethyl ether, petroleum ether Mineral oil and fixed vegetable oils Dielectric Constant of Solvent, ε (Approximately) 80 50 30 20 5 0 Decreasing Polarity ↓ Chapter 10| Solubility and Distribution The dielectric constant (Ԑ) of a compound is an index of its polarity which indicates the ability of solvent to separate two oppositely charged ions. A series of solvents of increasing polarity will show a similar increase in dielectric constant. Solubility depends on chemical, electrical & structural effects that lead to interactions between the solute and the solvent. The selection of the most suitable solvent is based on the principle of “like dissolves like”. That is, a solute dissolves best in a solvent with similar chemical properties. i.e. Polar solutes dissolve in polar solvents. E.g salts & sugar dissolve in water. Non polar solutes dissolve in non polar solvents. E.g. naphthalene dissolves in benzene. Chapter 10| Solubility and Distribution Classification of solvents& their mechanism of action: 1-Polar, 2- nonpolar, and 3- semipolar solvents 1. Polar solvents : Polar solvents (Water, glycols, methyl & ethyl alcohol), dissolve ionic solutes & other polar substances. Solubility of substances in polar solvents depends on structural features: 1. The ratio of the polar to the nonpolar groups of the molecule 2. Straight chain monohydroxy alcohols, aldehydes & ketones with >> 5 C are slightly soluble in water. Chapter 10| Solubility and Distribution 3. Branching of the carbon chain in aliphatic alcohols increases water solubility. Tertiary butyl alcohol >> soluble than n-butyl alcohol 4. Polyhydroxy compounds as glycerin, tartaric acid, PEG are water soluble (additional polar groups are present in the molecule). 1. Polar solvent acts as a solvent according to the following mechanisms: A. Dielectric constant: due to their high dielectric constant, polar solvents reduce the force of attraction between oppositely charged ions in crystals. Example: water possessing a high dielectric constant (> = 80) can dissolve NaCl, while chloroform (> = 5) & benzene (> = 2) cannot. Ionic compounds are practically insoluble in these 2 solvents. Chapter 10| Solubility and Distribution B. Solvation through dipole interaction: Polar solvents are capable of solvating molecules & ions through dipole interaction forces. The solute must be polar to compete for the bonds of the already associated solvent molecules. Example: Ion-dipole interaction between sodium salt of oleic acid & water Chapter 10| Solubility and Distribution C. Hydrogen bond formation: Water dissolves phenols, alcohols and other oxygen & nitrogen containing compounds that can form hydrogen bonds with water. Chapter 10| Solubility and Distribution D. Acid-base reaction: Polar solvents break covalent bonds of strong electrolyte by acid-base reaction because these solvents are amphiprotic HCl + H2O ⟶ H3O+ + Cl Chapter 10| Solubility and Distribution 2. Non polar solvents Non polar solvents such as hydrocarbon are: 1. Unable to reduce the attraction between the ions due to their low dielectric constants. 2. They are unable to form hydrogen bonds with non electrolytes. 3. Cannot break the covalent bond 4. Non polar solvents can dissolve non polar solutes through weak van der Waals forces Example: solutions of oils & fats in carbon tetrachloride or benzene. Chapter 10| Solubility and Distribution 3. Semipolar solvents Semipolar solvents, such as ketones can induce a certain degree of polarity in non polar solvent molecules. They can act as intermediate solvents to bring about miscibility of polar & non polar liquids. Example: acetone increases solubility of ether in water. Types of solutions Solutions of pharmaceutical importance include: I. Gases in liquids II. Liquids in liquids III. Solids in liquids Chapter 10| Solubility and Distribution I-Solubility of gases in liquids Examples of pharmaceutical solutions of gases include: HCl, ammonia water & effervescent preparations containing CO2 maintained in solution under pressure. The solubility of a gas in a liquid is the concentration of dissolved gas when it is in equilibrium with some of the pure gas above the solution. The solubility depends on the pressure, temperature, presence of salts & chemical reactions that sometimes the gas undergoes with the solvent Chapter 10| Solubility and Distribution 1. Effect of pressure According to Henry’s law: In a very dilute solution at constant temperature, the concentration (C2) of dissolved gas is proportional to the partial pressure (p) of the gas above the solution at Equilibrium. (The partial pressure of the gas = total pressure above the solution minus the vapor pressure of the solvent) Chapter 10| Solubility and Distribution C2 α p C2= σ p where C2 is the concentration of dissolved gas in gram/l of solvent, p is the partial pressure of the undissolved gas above the solution, σ is proportionality constant (solubility coefficient) Note: When the pressure above the solution is released (decreases), the solubility of the gas decreases, and the gas may escape from the container with violence. This phenomenon occurs in effervescent solutions when the stopper of the container is removed. Chapter 10| Solubility and Distribution Chapter 10| Solubility and Distribution Example A. if 0.016 g O2 dissolves in 1 liter of water at 25°C and at O2 pressure of 300 mmHg, calculate the solubility coefficient 𝐶2 = 𝜎𝑝 𝐶2 0.016𝑔/𝑙 𝜎= = 𝑝 300𝑚𝑚𝐻𝑔 = 5.33 × 10−5 ( 𝑔/𝑙)/𝑚𝑚𝐻𝑔 Chapter 10| Solubility and Distribution Example B. How many grams of O2 can be dissolved in 250ml of aqueous solution when the total pressure above the mixture is 760 mmHg? The partial pressure above the O2 in solution is 0.263 atm and the temperature is 25° 1 atm = 760mmHg, so 0.263 atm × 760 = 199.88mmHg 𝐶2 = 𝜎𝑝 𝐶2 = 5.33 × 10−5 ( g/l)/mmHg × (0.263 × 760)mmHg = 0.0107 g/l = 0.0027 g/250ml Chapter 10| Solubility and Distribution 2. Effect of temperature As the temperature increases the solubility of gases decreases, owing to the great tendency of the gas to expand Pharmaceutical application: The pharmacist should be cautious in opening containers of gaseous solutions in warm climates. A container filled with a gaseous solution or a liquid with high vapor pressure, such as ethyl nitrite, should be immersed in ice or cold water, before opening the container, to reduce the temperature and pressure of the gas. Chapter 10| Solubility and Distribution 3. Effect of Salting out : Adding electrolytes (NaCl) & sometimes non electrolytes (sucrose) to gaseous solutions (eg. carbonated solutions) induces liberation of gases from the solutions. Why? Due to the attraction of the salt ions or the highly polar electrolyte for the water molecules and reduction of the aqueous environment adjacent to the gas molecules. Chapter 10| Solubility and Distribution II- Solubility of liquids in liquids Preparation of pharmaceutical solutions involves mixing of 2 or more liquids (alcohol & water to form hydroalcoholic solutions, volatile oils & water to form aromatic waters, volatile oils & alcohols to form spirits …) Ideal and Real Solutions Chapter 10| Solubility and Distribution Ideal and Real Solutions: Ideal solution when the components of solution obey Raoult's law (adhesive forces =cohesive forces). Real Solutions when the components of solution not obey Raoult's law and are of two types: A. Negative deviated (adhesive forces >>cohesive forces) Negative deviations lead to increased solubility B. Positive deviations (cohesive forces >>cohesive forces) Positive deviations, leading to decreased solubility Chapter 10| Solubility and Distribution The attractive cohesive forces, which may occur in gases, liquids, or solids, are called internal pressures. Liquid-liquid systems may be divided into 2 categories: 1. Systems showing complete miscibility such as alcohol & water, glycerin & alcohol, benzene & carbon tetrachloride. 2. Systems showing Partial miscibility as phenol and water; two liquid layers are formed each containing some of the other liquid in the dissolved state. Complete miscibility occurs when: The adhesive forces between different molecules (A-B) >>cohesive forces between like molecules (A-A or B-B). Chapter 10| Solubility and Distribution Partial miscibility results when: Cohesive forces of the constituents of a mixture are quite different, e.g. water (A) and hexane (B). A-A » B-B The non polar molecules (B) will be squeezed out by the powerful attractive forces existing between the molecules of the polar liquid. The term miscibility refers to the mutual solubility of the components in liquid-liquid systems. Chapter 10| Solubility and Distribution Influence of Foreign Substances : If the added material is soluble in only one of the two components, the mutual solubility of the liquid pair is decreased. Example, if naphthalene is added to a mixture of phenol and water, it dissolves only in the phenol, the miscibility is decreased. If potassium chloride is added to a phenol-water mixture, it dissolves only in water and decreases the miscibility. If the added material is soluble in both of the liquids, the mutual solubility of the liquid pair is increased. Example, The addition of succinic acid or sodium oleate to a phenolwater system increases the mutual solubility. Chapter 10| Solubility and Distribution The increase in mutual solubility of two partially miscible solvents by another agent is ordinarily referred to as blending. SOLUBILITY OF SOLIDS IN LIQUIDS : Systems of solids in liquids include the most important type of pharmaceutical solutions. There is ideal and real (non ideal) solution of solids Chapter 10| Solubility and Distribution Ideal solution of solids : The solubility of a solid in an ideal solution depends on Temperature, (direct relationship) Melting point of the solid, (inverse relationship) Ideal solubility is not affected by the nature of the solvent. Molar heat of fusion, ∆ Hf, (The heat absorbed when the solid melts). In an ideal solution the heat of solution is equal to the heat of fusion, which is assumed to be a constant independent of the temperature. Chapter 10| Solubility and Distribution The equation derived from thermodynamic considerations for an ideal solution of a solid in a liquid is 𝜟𝑯𝒇 𝑻𝟎 − 𝑻 𝒊 𝐥𝐨𝐠 𝝌 = 𝟐 𝟐. 𝟑𝟎𝟑𝑹 𝑻𝑻𝟎 𝑖 where; 𝜒 is the ideal solubility of the solute expressed in mole 2 fraction, To is the melting point of the solid solute in absolute degrees, and T is the absolute temperature of the solution. At temperatures above the melting point, the solute is in the liquid state, and, in an ideal solution, the liquid solute is miscible in all proportions with the solvent. Therefore, the above equation no longer applies when T > To Chapter 10| Solubility and Distribution EXAMPLE 10-7: What is the solubility of naphthalene at 20°C in an ideal solution? The melting point of naphthalene is 80°C and the molar heat of fusion is 4500 cal/mole. R=1.987 Cal/mole.k 4500 353 − 293 𝑖 log 𝜒 = 2 2.303 × 1.987 293 × 353 𝑖 𝜒 = 0.27 2 Chapter 10| Solubility and Distribution Homework: calculate the solubility at 10 and 75 °C? You will see that the solubility increases as the temperature increased Chapter 10| Solubility and Distribution Q12. The m.p and molar heat of fusion of three indomethacin polymorphs I, II, VII are: polymorph m.p °C(K) ∆Hf cal/mole 𝑖 𝜒 2 I 158(431) 9550 0.0069 II 153(426) 9700 0.0073 VII 95(368) 2340 0.4716 Chapter 10| Solubility and Distribution Calculate the ideal mole fraction solubility at 25 °C (298) three indomethacin polymorphs and rank the solubility in decreasing order, is the m.p or ∆Hf more useful in ordering the solubility. According to m.p I>II>VII According to ∆Hf II>I>VII 𝜟𝑯𝒇 𝑻𝟎 − 𝑻 𝒊 𝐥𝐨𝐠 𝝌 = 𝟐 𝟐. 𝟑𝟎𝟑𝑹 𝑻𝑻𝟎 Chapter 10| Solubility and Distribution For I: 𝟗𝟕𝟎𝟎 𝒊 − 𝐥𝐨𝐠 𝝌 = 𝟐.𝟑𝟎𝟑×𝟏.𝟗𝟖𝟕 𝟐 𝟒𝟑𝟏−𝟐𝟗𝟖 𝟐𝟗𝟖×𝟒𝟑𝟏 𝒊 𝝌 = 𝟎. 𝟎𝟎𝟔𝟗 𝟐 For II: 𝟗𝟕𝟎𝟎 𝒊 − 𝐥𝐨𝐠 𝝌 = 𝟐.𝟑𝟎𝟑×𝟏.𝟗𝟖𝟕 𝟐 𝟒𝟐𝟔−𝟐𝟗𝟖 𝟐𝟗𝟖×𝟒𝟐𝟔 𝒊 𝝌 = 𝟎. 𝟎𝟎𝟕𝟑 𝟐 For VII: 𝟐𝟑𝟒𝟎 𝒊 − 𝐥𝐨𝐠 𝝌 = 𝟐.𝟑𝟎𝟑×𝟏.𝟗𝟖𝟕 𝟐 𝟑𝟔𝟖−𝟐𝟗𝟖 𝟐𝟗𝟖×𝟑𝟔𝟖 𝒊 𝝌 = 𝟎. 𝟒𝟕𝟏𝟔 𝟐 According to the solubility VII>II>I, so as the m.p increases, the solubility decreased Homework Q 13, 14, and 16 Chapter 10| Solubility and Distribution Non ideal Solutions : In non ideal solutions, the electrostatic and intermolecular forces should be considered. The activity of a solute in a solution is expressed as the concentration multiplied by the activity coefficient. When the concentration is given in mole fraction, the activity is expressed as a2 = X2ɣ2 where ɣ2 on the mole fraction scale is known as the rational activity coefficient. Converting to logarithms, we have log a 2= log X2 + log ɣ 2 Chapter 10| Solubility and Distribution 𝒊 In an ideal solution, a2 =𝝌 because ɣ 2 = 1, and, accordingly, the ideal 𝟐 solubility equation can be expressed in terms of activity as : 𝜟𝑯𝒇 𝑻𝟎 − 𝑻 𝒊 𝐥𝐨𝐠 a 𝟐 = −𝐥𝐨𝐠𝝌 = 𝟐 𝟐. 𝟑𝟎𝟑𝑹 𝑻𝑻𝟎 By combining the 2 equations, we find that the mole fraction solubility of a solute in a non ideal solution expressed in log form, becomes 𝜟𝑯𝒇 𝑻𝟎 − 𝑻 𝒍𝒐𝒈 𝑿𝟐 = + 𝒍𝒐𝒈 ɣ𝟐 𝟐. 𝟑𝟎𝟑𝑹 𝑻𝑻𝟎 Chapter 10| Solubility and Distribution Therefore, the mole fraction solubility in various solvents can be expressed as the sum of two terms: the solubility in an ideal solution and the logarithm of the activity coefficient of the solute. As a real solution becomes more ideal, ɣ 2 approaches unity, the equation returns 𝜟𝑯𝒇 𝑻𝟎 − 𝑻 𝒊 𝐥𝐨𝐠 𝝌 = 𝟐 𝟐. 𝟑𝟎𝟑𝑹 𝑻𝑻𝟎 Chapter 10| Solubility and Distribution Q18. the mole fraction solubility of naphthalene (nonpolar solute) in different solvents at temperature 40 °C (313 k), m.p 80°C(353k), ∆Hf =4500 14 cal/mole. Calculate the ɣ2 and find the relationship between X2 and ɣ2. X2 for chlorobenzen(nonpolar solvent)= 0.444, for water = 1.76 x10-5 𝜟𝑯𝒇 𝑻𝟎 − 𝑻 𝒍𝒐𝒈 𝑿𝟐 = + 𝒍𝒐𝒈 ɣ𝟐 𝟐. 𝟑𝟎𝟑𝑹 𝑻𝑻𝟎 For chlorobenzen log 0.444 = For water log 1.76 × 10 −5 4500 353 − 313 + log 𝛾 2 2.303𝑥1.987 313𝑥353 𝛾2 = 0.99 4500 353 − 313 = + log 𝛾2 2.303 × 1.987 313𝑥353 𝛾2 = 25003 So there is inverse relationship between X2 and ɣ2 Solubility of Strong Electrolytes Chapter 10| Solubility and Distribution Solubility of Strong Electrolytes : The effect of temperature on the solubility of some salts in water is shown in Figure 10-6. A rise in temperature increases the solubility of a solid that absorbs heat (endothermic process) when it dissolves. Such as KBr Conversely, if the solution process is exothermic, that is, if heat is evolved, the temperature of the solution rises and the container feels warm to the touch. Chapter 10| Solubility and Distribution The solubility in this case decreases with an elevation of the temperature. such as cerium(III) sulfate Most solids belong to the class of compounds that absorb heat when they dissolve Sodium sulfate exists in the hydrated form. Na2SO4.10H2O, up to a temperature of about 32°C, the solution process (dissolution) is endothermic, and solubility increase with temperature. Above this point, the compound exists as the anhydrous salt, Na2SO4, the dissolution is exothermic, and solubility decreases with an increase of temperature. Sodium chloride does not absorb or evolve an appreciable amount of heat when it dissolves in water; thus, its solubility is not altered much by a change of temperature, and the heat of solution is approximately zero. Chapter 10| Solubility and Distribution Chapter 10| Solubility and Distribution Chapter 10| Solubility and Distribution Solubility of Weak Electrolytes Many important drugs belong to the class of weak acids and bases. Weak acids react with dilute alkalies to form water-soluble salts, but they can be precipitated as the free acids if stronger acidic substances are added to the solution. For example, a 1% solution of phenobarbital sodium is soluble at pH values high in the alkaline range. The soluble ionic form is converted into molecular phenobarbital as the pH is lowered, and below 9.3, the drug begins to precipitate from solution. Chapter 10| Solubility and Distribution Weak bases react with dilute acids (decrease the pH) to form water-soluble salts, but they can be precipitated as the free bases if stronger basic substances (increase the pH) are added to the solution. For example, alkaloidal salts such as atropine sulfate begin to precipitate as the pH is elevated. To ensure a clear homogeneous solution and maximum therapeutic effectiveness, the preparations should be adjusted to an optimum pH. Chapter 10| Solubility and Distribution Calculating the Solubility of Weak Electrolytes as Influenced by pH : According to the Henderson-Hasselbach equation, the relationship between pH, pKa, and relative concentrations of an acid and its salt is as follows: Chapter 10| Solubility and Distribution where [A- ] is the molar concentration of the salt (dissociated species) and [HA] is the concentration of the undissociated acid. When the concentrations of salt and acid are equal, the pH of the system equals the pKa of the acid. As the pH decreases, the concentration of the molecular acid increases and that of the salt decreases. Changes in solubility brought about by alterations of solvent pH can be predicted by the pHp equation. Chapter 10| Solubility and Distribution The pHp is the pH below which an acid or above which a base will begin to precipitate. where, So = the molar solubility of the undissociated acid or base S = the molar concentration of the salt form of the drug initially added Chapter 10| Solubility and Distribution Example 10.16 : Below what pH will free phenobarbital begin to separate from a solution having an initial concentration of 1 g of sodium phenobarbital per 100 mL at 25°C? solution : The molar solubility, So, of phenobarbital is 0.0050 and the pKa is 7.41 at 25°C. The molecular weight of sodium phenobarbital is 254. The molar concentration of salt initially added is (g/liter)/m.wt = 10/254= 0.039mole/liter pHp= 7.41+log 0.039−0.005 0.005 =8.24 Chapter 10| Solubility and Distribution Q38. The molar solubility of sulfathiazole (weak acid) in water is 0.002, the pka=7.12, m.wt of sodium sulfathiazole = 304, what is the lowest pH allowable for complete solubility in a 5% solution of salt? 𝑤𝑡 1000 M salt = × 𝑚 ⋅ 𝑤𝑡 𝑣𝑜𝑙 5 1000 = × = 0.164 304 100 (0.164 − 0.002) pHp = 7.12 + log = 9.03 0.002 Homework Q 39 and 40 Chapter 10| Solubility and Distribution Solubility of Slightly Soluble Electrolytes : The Influence of Solvents on the Solubility of Drugs Weak electrolytes can behave like strong electrolytes or like nonelectrolytes in solution When the solution is of such a pH that the drug is entirely in the ionic form, it behaves as a solution of a strong electrolyte--------> no problem. Chapter 10| Solubility and Distribution However, when the pH is adjusted to a value at which un-ionized molecules are produced in sufficient concentration to exceed the solubility of this form, precipitation occurs. --------> (Problem). To solve this problem, a solute is more soluble in a mixture of solvents than in one solvent alone. This phenomenon is known as cosolvency, and the solvents that, in combination, increase the solubility of the solute are called cosolvents. For example phenobarbital solubility is increased when add alcohol or glycerin to water. Chapter 10| Solubility and Distribution Combined Effect of pH and Solvents The solvent affects the solubility of a weak electrolyte in a buffered solution in two ways: The addition of alcohol to a buffered aqueous solution of a weak electrolyte increases the solubility of the un-ionized species by adjusting the polarity of the solvent to a more favorable value. Because it is less polar than water, alcohol decreases the dissociation of a weak electrolyte, and the solubility of the drug goes down as the dissociation constant is decreased (pKa is increased). Chapter 10| Solubility and Distribution Influence of Surfactants Weakly acidic and basic drugs can he brought into solution by the solubilizing action of surface-active agents such as detergent. Influence of complexation Complexation may be used to increase the solubility such as addition of potassium iodide to iodine. Sometimes complexation cause decrease in solubility such complex between tetracycline and calcium produce insoluble complex Chapter 10| Solubility and Distribution Influence of size and shape of particles The size particles affect solubility. Solubility increases with decreasing particle size as consequence of increase in surface area. The configuration of a molecule and the type of arrangement in the crystal also has some influence on solubility, and a symmetric particle can be less soluble than an unsymmetric one. Chapter 10| Solubility and Distribution Solubility of Slightly Soluble Electrolytes : When slightly soluble electrolytes are dissolved to form saturated solutions, the solubility is described by a special constant, known as the solubility product, Ksp , of the compound. Silver chloride is an example of such a slightly soluble salt. The excess solid in equilibrium with the ions in saturated solution at a specific temperature is represented by the equation Chapter 10| Solubility and Distribution and because the salt dissolves only with difficulty and the ionic strength is low, the equilibrium expression can be written in terms of concentrations instead of activities: [Ag+][Cl− ] =K [AgCl solid] Moreover, because the concentration of the solid phase is essentially constant, [Ag+][Cl- ]= Ksp Chapter 10| Solubility and Distribution As in the case of other equilibrium expressions, the concentration of each ion is raised to a power equal to the number of ions appearing in the formula. Thus, for aluminum hydroxide, Al(OH)3. [A13+][OH-]3 = Ksp Chapter 10| Solubility and Distribution EXAMPLE 10-13: The measured solubility of silver chloride in water at 20°C is 1.12 x 10-5 mole/liter. This is also the concentration of the silver ion and the chloride ion because silver chloride is nearly completely dissociated. Calculate the solubility product of this salt. We have Ksp = (1.12 x 10-5) x (1.12 x 10-5) = 1.25 x 10-10 If an ion in common with AgCI, that is, Ag+ or Cl-, is added to a solution of silver chloride, the equilibrium is altered. Chapter 10| Solubility and Distribution The addition of sodium chloride, for example, increases the concentration of chloride ions so that momentarily [Ag+][Cl- ] > Ksp and some of the AgCl precipitates from the solution (the reaction shift to left) until the equilibrium [Ag+][Cl-] = Ksp is reestablished. Hence, the result of adding a common ion is to reduce the solubility of a slightly soluble electrolyte Salts having no ion in common with the slightly soluble electrolyte produce an effect opposite to that of a common ion: At moderate concentration, they increase rather than decrease the solubility because they lower the activity coefficient. Chapter 10| Solubility and Distribution Distribution of Solutes between Immiscible Solvents If an excess of substance 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. Chapter 10| Solubility and Distribution If C1 and C2 are the equilibrium concentrations of the substance in Solvent1 and Solvent2, respectively, the equilibrium expression becomes 𝒄𝟏 K= 𝒄𝟐 The above equation is known as the distribution law. The equilibrium constant, K, is known as the distribution ratio, distribution coefficient, or partition coefficient. Chapter 10| Solubility and Distribution The partition law states that: At a given temperature, the ratio of the concentrations of a solute in two immiscible solvents (solvent 1 and solvent 2) is constant when equilibrium has been reached This constant is known as the partition coefficient (or distribution coefficient) Chapter 10| Solubility and Distribution Example 9-5 :When boric acid is distributed between water and amyl alcohol at 25°C, the concentration in water is found to be 0.0510 mole/liter and in amyl alcohol it is found to be 0.0155 mole/liter. What is the distribution coefficient? We have 𝐂 ⋅ 𝐻2𝑜 0.051 K= = = 3.29 𝐂 ⋅ alc 0.0155 Chapter 10| Solubility and Distribution No convention has been established with regard to whether the concentration in the water phase or that in the organic phase should be placed in the numerator. Therefore, the result can also be expressed as 𝐂 ⋅ alc 0.0155 K= = = 0.304 𝐂 ⋅ 𝐻2𝑜 0.0510 One should always specify, which of these two ways the distribution constant is being expressed. Chapter 10| Solubility and Distribution Importance of partition Knowledge The principle is involved in several areas: 1. Drugs partitioning between aqueous phases and lipid biophases 2. Preservation of oil–water systems 3. Absorption and distribution of drugs throughout the body 4. Antibiotics partitioning into microorganisms 5. Solvent extraction 6. Chromatography Chapter 10| Solubility and Distribution Partition law holds true 1. at constant temperature 2. when the solute exists in the same form in both solvents (the species are common to both phases) Chapter 10| Solubility and Distribution Effect of Ionic Dissociation and Molecular Association on Partition: Some compounds have more than one species in oil phase and in aqueous phase which produce complicated case in calculation of partition coefficient. Example, benzoic acid that is used as preservative present in oil phase as monomer and as dimer in equilibrium. Chapter 10| Solubility and Distribution Benzoic acid present in aqueous phase as unionized (HA) and as ionized (A- ) in equilibrium. The true distribution coefficient, K is the ratio of molar concentration of the species common to both the oil and water phases Chapter 10| Solubility and Distribution The experimentally observed or apparent distribution coefficient determined by using the total acid concentration (all species) obtained by analysis is. Thus the observed distribution coefficient depends on two equilibria. In the oil phase, the concentration of monomer or dimer depends on the type of oil, for example benzoic only present as monomer in peanut oil. In the aqueous phase, the concentration of unionized or ionized depends on the pKa of compound and pH of solvent. Chapter 10| Solubility and Distribution Extraction Chapter 10| Solubility and Distribution Extraction : Liquid-liquid extraction is a useful method to separate components of a mixture. Liquid-liquid extraction is based on the transfer of a solute substance from one liquid phase into another liquid phase according to the solubility. Chapter 10| Solubility and Distribution The success of this method depends upon the difference in solubility of a compound in various solvents. For a given compound, solubility differences between solvents is quantified as the "distribution coefficient" Example, suppose that you have a mixture of sugar in vegetable oil (it tastes sweet!) and you want to separate the sugar from the oil. You suspect that the sugar is partially dissolved in the vegetable oil. To separate the sugar from the oil we add water to the mixture with shaking. Chapter 10| Solubility and Distribution Sugar is much more soluble in water than in vegetable oil, and water is immiscible (=not soluble) with oil. By shaking the sugar will move to the phase in which it is most soluble: the water layer At the end, the water phase tastes sweet, because the sugar is moved to the water phase upon shaking. ((You extracted sugar from the oil with water. )) Chapter 10| Solubility and Distribution Chapter 10| Solubility and Distribution Example, Iodine can be extracted from water by adding hexane, shaking and separating the two layers in a separating funnel. To determine the efficiency with which one solvent can extract a compound from a second solvent we use the following equations: The distribution coefficient is K= 𝑾𝟏/𝑽𝟏 𝑾−𝑾𝟏 𝑽𝟐 Chapter 10| Solubility and Distribution By rearrangement, weight of solute extracted can be calculated using this equation: Wn=W 𝒏 𝑲𝑽𝟏 𝑲𝑽𝟏+𝑽𝟐 w = weight in grams of a solute is extracted repeatedly V1 = volume in mL of original solvent V2 = volume in mL of a second solvent (extraction solvent) w1 = weight of the solute remaining in the original solvent after extraction n= number of extraction repeating Note: It can be shown from the equation that a most efficient extraction result when n is large and V2 is small. Chapter 10| Solubility and Distribution Example 9-7 : The distribution coefficient for iodine between water and carbon tetrachloride at 25°C is K=CH2O/CCCl4 = 0.012. How many grams of iodine are extracted 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? We have 0.012 × 50 𝑊1 = 0.10 × (0.012 × 50) + 10 = 0.0057 𝑔 remains or 0.0943 𝑔 is extracted 0.012 × 50 𝑊2 = 0.10 × (0.012 × 50) + 5 2 = 0.0011 𝑔 of iodine Thus, 0.0011 g of iodine remains in the water phase, and the two portions of CCl4 have extracted 0.0989 g. Chapter 10| Solubility and Distribution To calculate the total concentration of benzoic acid that must be added to preserve an oil-water mixture, we can use the following equations: C = (Kq+1+ Ka/[H3O+])[HA]w where, C= total concentration of acid that must be added to the two-phase system to obtain a final specified concentration [HA]w of undissociated acid in the aqueous phase buffered at a definite pH or hydrogen ion concentration K= the distribution coefficient= [HA]o / [HA]w Chapter 10| Solubility and Distribution q = the volume ratio of the two phases, is needed when the volumes are not equal = Vo/Vw Ka= the dissociation constant of the acid in the aqueous phase By rearrangement the equation we can calculate the 𝐶 [HA]w= Kq+1+ Ka/[H3O+] Chapter 10| Solubility and Distribution EXAMPLE 10-25: If benzoic acid is distributed between equal volumes of peanut oil and water, what must be the original concentration in the water phase in order that 0.25 mg/mL of undissociated acid remains in the aqueous phase buffered at a pH of 4.0? The partition coefficient, K = [HA]o/[HA]W, is 5.33 and the dissociation constant of the acid in water is 6.4 x 10-5. Because the two phases are present in equal amounts, q = Vo/Vw= 1. C = (Kq+1+ Ka/[H3O+])[HA]w C= (5.33+1+ (6.4x10-5/10-4))0.25 = 1.74 mg/m