PCT 201 Physical Pharmacy I PDF

1 CONCENTRATION This refers to the amount of substance in another or the amount of a solute dissolve in a solvent. The concentration of solute is express in a number of different forms. Each of these con...

1 CONCENTRATION This refers to the amount of substance in another or the amount of a solute dissolve in a solvent. The concentration of solute is express in a number of different forms. Each of these concentration expressions is used in different applications in the Pharmaceutical sciences. It is important to state here that concentration should not be confused for dose as it is always the case with students in their early years. While concentration refers to the amount of substance in another or the amount of a solute dissolve in a solvent, a dose is the quantitative amount of a drug to be administered or taken by a patient for the intended medicinal effect. The lecture objective is for students to be able to define concentration and clearly understand each concentration expression. EXPRESSIONS OF CONCENTRATION The concentration of solution can be express either in terms of the quantity of solute in a definite volume of a solution or as the quantity of a solute in a definite mass of a solvent or solution. The various expressions of concentrations are: 1. Number of moles 2. The mole fraction 3. The mole per cent 4. Molarity 5. Molality 6. Normality 7. Osmolarity and Osmolality 8. Percentage by weight (% w/v) 9. Percentage by volume (% v/v) 10. Percentage weight in weight 11. Part per million (ppm) The Mole (n) The weight of a solute (in grams) divided by its molar mass converts the weight/mass of that solute to moles. mass Number of moles = molar mass PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  2 Example 1 How many moles of NaCl are in 58.4426g of the substance? [Na = 22.9898, Cl = 35.4528] Molar mass of NaCl = 22.9898 + 35.4528 = 58.4426 mass 58.4426g Number of moles = molar mass = =1 58.4426 Therefore 1 mole of NaCl is 58.4426g Example 2 1. How many moles of monobasic sodium phosphate are present in 100 g of the substance? [M.wt = 138] Answer; mass 100 Number of moles = molar mass = = 0.725 moles 138 Exercise 1. How many moles are in 17.7 g of KMnO4? 2. How many moles are in 41.2 g of Mg3(PO4)2? 3. How many moles are in 26.9 g of C12H22O11? (Get the atomic weights from any periodic table) The above examples indicates that the mole is a way of expressing concentration since it can be used in expressing the amount of a substance in another Mole fraction (X) It gives a measure of the relative proportion of moles of each constituent in a solution. For a system of two constituents (e.g. solute and solvent) n1 n2 X1 = n1+n2 X2 = n2+n1 X1 is the mole fraction of constituent 1 (solvent) X2 is the mole fraction of constituent 2 (solute) n1 = number of moles of the solvent n2 = number of moles of the solute Mole fraction is the number of moles of a particular constituent divided by the number of moles of all the constituents. The sum of the mole fractions of solute and solvent must equal 1. In a solution containing 0.01 mole of solute and 0.04 mole of solvent, the mole fraction of the solute is PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  3 0.01 X2 = = 0.20 0.04+0.01 Because the mole fraction of the two constituents must equal 1, the mole fraction of the solvent is 0.8. Mole fraction allows one to express the relationship between the number of solute and the number of solvent molecules in a system. In the above example, it can be readily seen that 2 of every 10 molecules in the solution are solute molecules. Mole Percent This is obtained by multiplying the mole fraction by 100. n1 n2 X1 = n1+n2 X 100 X2 = n2+n1 X 100 Molarity (M) This is the number of moles of solute contained in 1 litre of solution. number of moles Molarity = volume of solution in litre The unit of molarity is moldm-3 (1 litre = 1 dm3) 0.1 molar solution contain 0.1 mole of solute per litre of the solution and is written as 0.1M. Molality (m) This is the number of moles of solute in 1 kg of solvent. Moles of solute Molality = mass of solvent in Kg The S.I unit of molality = mol/Kg. Molal solutions are prepared by adding the proper weight of solvent to a carefully weight quantity of the solute. Note: Molarity involves volume of solution whereas molality involves mass of solvent. Molality has some advantages over molarity unit because molality is independent of temperature of the solution whereas molarity varies with temperature because volume of the solution varies with temperature. Normality (N) Normality is the gram equivalent weight of solute in 1 litre of solution. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  4 The S.I unit of normality is g/L. Molecular mass Gram equivalent weight = Valency Osmolarity and Osmolarity The amount of osmotically active particles in a solution is sometimes expressed in terms of osmoles. These particles may either be molecules or ions. Osmoles are the number of particles dissolved in a solution regardless of charge. Osmolarity is the number of osmoles per litre of solution. For substances that maintain their molecular structure when the dissolve (e.g. glucose), osmolarity and molarity are essentially the same. For substances that dissociate when the dissolve, the osmolarity is the number of free particles multiply by the molarity. Thus, for a 1 molar solution of pure NaCl solution, the osmolarity is; 2 x 1= 2. ie 2 osmolar (Molarity = 1, number of free particles = 2; ie 1 Na+, and 1 Cl-). Parenteral fluids labels are require to state the osmolar concentration of the preparation. Such information indicates to the user/ practioner whether the solutions are hypo-osmotic, iso- osmotic, or hyper-osmotic with regards to biologic fluids and membranes. The formula below is used in calculating osmolarity mass of substance (g/L) Osmolarity = X number of free particles molar mass Osmolality is the number of osmoles per kilogram of solvent. Percentage by weight (% w/w) This signifies the number of grams of solute in 100 g of solution. A 10% w/w aqueous solution of glycerine contains 10 g of glycerine in enough water (90 g) to make 100 g of solution. Percentage by volume (% v/v) It signifies the volume of solute in millilitres contained in 100 ml of the solution or liquid preparation. Percentage weight in volume (% w/v) It signifies the number of grams of solutes in 100 ml of solution or liquid preparation. Example What is the osmolarity of 0.9% (w/v) sodium chloride injection? Answer 0.9% = 0.9g in 100 ml Therefore, if 0.9 g → 100 ml PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  5 X g → 1000 ml (note that 1000ml = 1L) 0.9 X 1000 X= = 9 g/L 100 Number of free particles in NaCl = 2; ie 1 Na+, and 1 Cl-) mass of substance (glL) Therefore, Osmolarity = X number of free particles molar mass 9 = 58.5 x 2 = 0.308 osmol/L Part per million (ppm) This is a way of expressing very dilute concentrations of substances. Just as per cent means out of a hundred, so parts per million or ppm means out of a million. Usually describes the concentration of something in water or soil. For example, fluoridated drinking water, used to reduce dental caries, often contains 1 part of fluoride per million parts of drinking water (1:1,000,000). ppm concentration of a substance may be expressed in quantitatively equivalent values of percent strength or ratio strength Example 1. Express 5 ppm of iron in water in percent strength. Answer 5ppm = 5 g in 1,000,000 g Thus; if 5 g → 1,000,000 g x g → 100 g 100 𝑋 5 = 1,000,000 = 0.0005 % IDEAL AND REAL SOLUTION Raoult law If the two components of a solution are volatile, they will have a natural tendency of escaping into vapor phase and exerting a definite vapor pressure. When the solution is in equilibrium with vapor, the total pressure above the solution will be equal to the sum of the partial vapor pressures of the individual components in accordance with Dalton’s Law of partial pressures. If 𝑷𝑨 𝐚𝐧𝐝 𝑷𝑩 are the partial vapor pressures of two volatile and miscible components A and B, the total pressure, P, is; P = PA + PB - - - - - - - (1) P = Total vapour pressure above the solution PA = Partial vapour pressure exerted by the solute PB = Partial vapour pressure exerted by the solvent PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  6 The assumption made here is that vapors behave as ideal gases. The partial vapor pressure of each component of an “ideal solution” can be easily determined from Raoult’s Law. This law states that, the partial vapor pressure of any volatile constituent of a solution is equal to the vapor pressure of the pure constituent multiplied by the mole fraction of that constituent in solution over the whole concentration range at all temperatures. i.e. PA = PAo XA - - - - - - - (2) From equation (1) and (2) P = PA + PB = PAo XA + PBo XB - - - - - (3) PAo = vapour pressure exerted by component A PBo = vapour pressure exerted by component B If you have a solution whose total vapour pressure is the sum of the partial vapour pressure of the components that constitute it, then it is said to obey Raoult’s law. In other words, if the total vapour pressure of the solution is described by equation (3) then Raoult’s law is obeyed by the system. Ideal solutions An ideal solution may be defined as one that obeys Raoult’s law. Properties of an ideal solution 1. It is a solution in which there is complete uniformity of cohesive forces. 2. The components of ideal solution have a similar chemical structure and because of such similarity, the intermolecular force the components of the solution will be uniform. i.e., the molecules in an ideal solution will exert the same force on each other irrespective of their nature. Thus if we have an ideal solution of two components, A and B, the intermolecular forces between A and A, B and B, and A and B, will be same. 3. No heat is given out or absorbed in the formation of solution and 4. The total volume is the sum of the component volumes; the volume does not shrink or expand 5. In ideal solutions, there is no change in the physical properties of the components other than dilution when they are mixed. The relationship between the vapour pressure for a liquid and its mole fraction in solution is given by Raoults law; the vapour pressure of each volatile constituent is equal to the vapour pressure of the pure constituent multiplied by its mole fraction in the solution as shown in equation 1 to 3 above. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  7 Only a few actual solutions show ideal behaviour, examples of system that exhibit ideal behaviour are mixtures of the following; Benzene + Toluene n-Hexane + n – heptane Ethyl Bromide + ethyl iodide The figure below is the vapour pressure composition diagram for an ideal binary system Vapour pressure composition curve for an ideal binary system Real solutions As pointed out earlier, only a few binary solutions strictly obey Raoult’s law over the whole range of concentrations. In many cases, mixtures of two miscible liquids form non-ideal solutions over most of the concentration range and the systems exhibit either a positive or negative deviation from Raoult’s law. These deviations from Raoult’s law are generally observed for systems in which the two components A and B of the liquid mixture have different molecular structure. Due to differences in the molecular structure, intermolecular force will not be same. If the attraction between the different molecules, i.e., A and B molecules is weaker than between similar molecules i.e, between A-A and B-B molecules, the escaping tendency from the solution into the vapor of the molecules of each kind will be more than from the two pure liquids. The partial vapor pressures of A and B will, therefore, be greater than predicted PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  8 from Raoult’s law. Such a system will exhibit positive deviation from Raoult’s law. The shape of the curve is as shown below. Vapor pressure of solutions exhibiting positive deviations from Raoult’s law (The dotted lines show ideal behaviour) In this case, the system exhibits a maximum in the total vapor pressure curve. Examples of this type of system are: acetone-carbon disulphide, chloroform-ethyl alcohol, benzene-cyclohexane etc. Large positive deviations are observed in the case of mixtures of two liquids which markedly differ in polarity, length of hydrocarbon chain and degree of association e.g., an alcohol and a hydrocarbon. If the difference in polarity between the two liquids is quite small e.g., ether and acetone; or both the two liquids are non-polar e.g., carbon tetrachloride and heptane, the positive deviations from Raoult’s law are small. If the attraction between different molecules i.e., between A-B molecules is stronger than between identical molecules i.e., between A-A and B-B molecules, the escaping tendency from the solution into the vapor of molecules of each kind will be small than the two pure liquids. As a result of this, partial vapor pressure of each constituent will be smaller than that predicted from Raoult’s law and the system will exhibit negative deviation from Raoult’s law. The shape of the curve is as shown below PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  9 Vapor pressure of solutions exhibiting negative deviations from Raoult’s law (The dotted lines show ideal behaviour) In this case the system exhibits a minimum in the total vapour pressure curve, examples of this type of system are pyridine-acetic acid, , chloroform-acetone, water - nitric acid etc. Classification of real solutions The occurrence of deviation from ideal behaviour allows the classification of solutions of liquid into three types: 1. Systems where the total vapour pressure is always intermediate between those of the pure components. These are known as zeotropic mixtures. Examples Carbon tetrachloride + cyclohexane, Water + methanol 2. Systems that exhibit a maximum value in the vapour pressure composition diagram. They are known as azeotropic mixture with a maximum vapour pressure e.g. Benzene + ethanol Water + ethanol 2. Systems that exhibit a minimum value in the vapour pressure composition diagrams. They are known as azeotropic mixtures with a minimum vapour pressure e.g. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  10 Chloroform + acetone Pyridine + acetic acid Water + nitric acid Henry’s law Henry’s law is used for the study of gas solubility. Henry’s law states that; the mass of a gas dissolved by a given volume of liquid at constant temperature is directly proportional to the pressure of the gas. MαP M = KP m =K p M = mass of gas dissolved per unit volume of a solvent P = pressure of the gas in equilibrium with the solution K = constant. This law is strictly applicable only at low pressure and higher temperature. The higher the temperature and the lower the pressure, the more closely the law is obeyed. The law is not applicable when the dissolved gas react with the solvent or where the dissolved gas ionise e.g. the law is valid for the solubility of NH3 in benzene and not in water in which a part of ammonia react with water to form NH4OH (Ammonium hydroxide) which in turn dissociate partially to NH3 + H2O NH4OH NH4OH NH+4 + OH- DISTILLATION OF BINARY LIQUIDS The different behaviors of solutions of liquid in liquid mixtures as noted from Raoults law affects their distillation and is of importance in various pharmaceuticals fields. Fractional distillation is used for the separation of completely miscible mixture. The method is effective only in case of zeotropic mixtures. It fails in case of those mixtures whose vapor pressure composition curve show either maximum or minimum. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  11 Boiling point composition diagram for a zeotropic system Let the composition of a liquid mixture (of liquid X and Y) be represented by J. From the figure above, it is clear that this liquid mixture will have component X in excess. If the mixture is heated to temperature C where it start boiling (meaning C is the boiling point) of the mixture), at this temperature the composition of the vapor is in equilibrium with the mixture as indicated by E. If this vapor is condensed, the distillate will have the composition represented by G, thus the distillate will be richer in Y than original liquid, because Y has a lower boiling point and hence more volatile. If the distillate (G) so obtained is redistilled, it will boil at a temperature F, which is below C. At this temperature, the composition of the vapors is given by K and the corresponding composition of the distillate will also be K, which is very close to pure Y, in this way, by repeating the fractional distillation process with the distillate we can get pure Y and effect the complete separation of the mixture into its component. Azeotropes with maximum vapour pressure and minimum boiling point When liquids in a solution do not have great chemical affinity for each other (positive deviation from ideality) their greater escapes tendencies increases the vapour pressure much more than expected on the basis of Raoults law. Thus, the boiling point of such an azeotrope is less than the boiling temperature of any of the constituents. Such azeotrope are either called positive azeotrope, minimum boiling mixture or pressure maximum azeotropes. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  12 Phase diagram of a positive azeotrope It will be found that complete separation into pure X and pure Y cannot be achieved. Point A is the boiling point of a known azeotropic mixture. The vapour that seperates at that temperature has composition B. The vapour at B is richer in constituent X than the liquid at point A. the vapour is cooled at point C, where it condenses. If the collected liquid (point C) is boiled again, it progresses to point D and so on. The stepwise progression shows how repeated distillation can never produce a distillate that is richer in constituents X than the azeotrope. Note that starting from the right of the azeotrope result in the same step wise process closing in on the azeotrope point. Azeotrope with minimum vapour pressure and maximum boiling point When liquid solutions form chemical bond (negative deviation from ideality) their escape tendencies and hence the vapour pressure decreases than expected on the basis of Raoults law. Thus the boiling point of such an azeotrope is greater than the boiling point of any of it constituent. Such azeotropes are either called negative azeotrope, maximum boiling mixtures or pressure minimum azeotropes. The point A, shown in the diagram below is a boiling point with a composition chosen very near to the azeotrope. The vapour is collected at the same temperature at point B. the vapour is PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  13 cooled, condensed and collected at point C. The distillate is poorer in constituents X and richer in constituents Y than the original mixture. Because this process has removed a greater fraction of Y from the liquid than it had originally, the residue must be poorer in Y and richer in X after distillation than before. No amount of distillation can make either the distillate or the residue arrive on the opposite side of the azeotrope from the original mixture. Phase diagram of a negative azeotrope COLLIGATIVE PROPERTIES It has been observed that any solution containing some non-volatile solutes exhibits these four properties; namely,  Vapor pressure lowering of the solvent.  Boiling point elevation of the solution.  Freezing point depression of the solution.  Osmosis and osmotic pressure of the solution. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  14 These four properties are collectively known as the colligative properties of the solution. A colligative property is that property which depends only upon the number of solute particles present in the solution and in no way depends upon the chemical nature of these particles. The solute present should not undergo association or dissociation. If there is association of solute particles, the number of particles actually added will become less and correspondingly, the properties mentioned above will be lowered. Similarly, if the solute particles are dissociated as in the case of electrolytes, the number of particles becomes larger than the original number and the value of the colligative properties become higher. Therefore, for such solutions in which the solute particles undergo association, or dissociation, some modification is essential over the laws deduced for solutions in which the solute particles are not associated or dissociated. While discussing these properties, it will be assumed that the solutions are very dilute and are ideal i.e. the interactions between the solute particles are negligibly small. Osmotic pressure Osmosis is the spontaneous diffusion of solvent from a solution of low solute concentration (or a pure solvent) into a more concentrated one through a semi permeable membrane osmotic pressure of a solution is the external pressure that must be apply to the solution in order to stop it from being diluted by the entry of solvent via osmotic flow. The semi-permeable membrane separates the two solutions and is permeable only to solvent molecules. The driving force for osmosis arises from the inequality of the chemical potential of the solvents on opposing sides of the membrane. Thus the direction of osmotic flow is from dilute solution (or pure solvent), where the chemical potential of the solvent is highest because of higher concentration of solvent molecules into the concentrated solution where the chemical potential of the solvent are reduced by the presence of more solute. The pressure that must be applied to a more concentrated solution to prevent the flow of pure solvent into the solution separated by a semi permeable membrane is called osmotic pressure. The relationship The relationship between osmotic pressure (P) and concentration for a non-electrolyte is given for by the Vant Hoff equation; Pv = n2RT v = volume of solution n2 = no. of moles of solute R = gas constant PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  15 T = Temperature One of the most important pharmaceutical applications of osmotic pressure is in the preparation of isotonic intravenous solution. Solutions that have the same or equal osmotic pressure with blood are called isotonic solutions. Those that are more dilute than isotonic are called hypotonic, and those that are more concentrated are called hypertonic. Intravenous solutions that are isotonic with blood include 0.9% NaCl and 5.5% dextrose. Solutions that are isotonic with lacrimal fluids (fluids from the lacrimal gland e.g Tears) include 0.9% NaCl and 1.9% boric acid. Red blood cells (erythrocytes) have been studied by immersion into solution of varying concentration of difference solutes. When RBC are introduced into water or into NaCl solutions containing less than 0.9g of NaCl per 100ml (hypotonic solution), the RBC swelled and often burst, this is refer to as haemolysis. If the cells are placed in solutions containing more than 0.9g of NaCl in 100.ml (hypertonic solution), the lose water and shrink. But when they are immersed in a solution containing exactly 0.9g of NaCl in 100ml (isotonic solution), no change in size of the cell is observed, because in such solution the cell maintain their tone, the solution is said to be isotonic with the human erythrocytes. Therefore, it is desirable that solutions to be injected into the blood be made isotonic with erythrocytes. Also, solutions which are not isotonic will cause pain at the site of injection and post infusion phlebitis (inflammation of a vein), tissue damage and necrosis. Vapour pressure lowering When a non-volatile solute is dissolved in a liquid solvent, it will alter/lower the tendency of the liquid to escape from the original liquid. In an ideal solution or one that is very dilute, the partial vapour pressure (P1) of one component is proportional to the mole fraction of molecules (X1) of that component in the mixture: P1 = P1o X1 P1o = vapour pressure of the pure solvent Example What is the partial vapour pressure of a solution containing 50 g dextrose in 1000 mL of water? The vapour pressure of water is given as 23.76 mmHg. Molar mass of dextrose is 180 50 g Number of mole of dextrose = = 0.28 mol 180 To calculate the number of moles of water, we need to first calculate the mass of water, Recall, Density of water = 1, volume of water = 1000 mL PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  16 mass Density = , mass = density x volume = 1 x 1000 g volume Number of moles of water = 0.28 + 55.56 = 55.84 moles 55.56 Mole fraction of water = 0.28+55.56 = 0.995 P1 = 0.995 x 23.76 mmHg = 23.64 mmHg The vapour pressure of the solution is 23.64 mmHg. The decrease in vapour pressure by addition the 50 g dextrose is 23.76 – 23.64 = 0.12mmHg Boiling Point Elevation The boiling point of a liquid is the temperature at which the vapour pressure of the liquid comes into equilibrium with the atmospheric pressure. The vapour pressure is reduced when a non volatile solute is added to a solvent, so the solution must reach a higher temperature to re- establish the equilibrium, hence, an increase in the boiling point. This is described in the equation: ΔTb = KbM Where, ΔTb = change in boiling point Kb = molal elevation constant of water M = molality Example What is the boiling point elevation of a solution containing 50 g of dextrose in 1000 mL of water? The molal elevation constant of water is 0.51. 50 g Number of mole of dextrose = = 0.28 moles of dextrose in 1000 mL of water. 180 Recall, Density of water = 1, volume of water = 1000 mL mass Density = volume, mass = density x volume = 1 x 1000 g Mass = 1000g = 1Kg Moles of solute 0.28 Molality = = = 0.28 moles/Kg mass of solvent in Kg 1 ΔTb = 0.51 x 0.28 = 0.143 oC Freezing point depression The freezing point of a pure liquid is the temperature at which the solid and liquid phases are in equilibrium at 1 atmosphere. The freezing point of a solution is the temperature at which the solid phase of pure solvent and the liquid phase of solution are in equilibrium at 1 atmosphere PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  17 pressure. When a solute is added to a solvent, the decrease in freezing point is proportional to the concentration of the solute. The relationship is described by the following equation: ΔTf =Kfm Where, ΔTf = freezing point elevation Kf = molal freezing point depression constant of water M = molality of solute (moles of solute per Kg of solvent) Example What is the decrease in freezing point of a solution containing 50 g of dextrose in 1000 mL of water? The molal freezing point depression constant of water is -1.86 oC. 50 g Number of mole of dextrose = = 0.28 moles of dextrose in 1000 mL of water. 180 Recall, Density of water = 1, volume of water = 1000 mL mass Density = volume, mass = density x volume = 1 x 1000 g Mass = 1000g = 1Kg Moles of solute 0.28 Molality = = = 0.28 moles/Kg mass of solvent in Kg 1 ΔTf = -1.86 x 0.28 = -0.52 oC Colligative Properties of Solutions of Electrolytes While discussing the four properties, it was assumed that solute persists in the same molecular form in the solvent. Such solutions obey the various relations deduced there. On the other hand, there are some solutes which associate or dissociate in the solvent and yield abnormal results when the various relations deduced are applied to them. Consider solutions of electrolytes such as aqueous solutions of strong acids, bases and their salts. In such solutions, the solute has a tendency to dissociate into ions, thereby, increasing the number of particles in the solutions. Consequently for such solutions, the freezing point lowering, boiling point elevation, vapor pressure lowering and osmotic pressure are higher than for non-electrolytes when solutions having the same concentration of electrolytes and non-electrolytes are compared. In order to account for all abnormal cases and to represent colligative properties of electrolytes by means of relations for non-electrolytes, Van’t Hoff introduced the factor i, known as the Van’t Hoff factor. This factor is defined as the ratio of the colligative effect produced by a PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  18 definite concentration of the electrolyte divided by the effect observed for the same concentration of a non-electrolyte. That is; observed colligative properties i= colligative properties expected if dissociation did not occur The value of i can be calculated from experimental data for each electrolyte at various concentrations. The value, calculated for a particular concentration of an electrolyte for one of the colligative properties, is found to be same and valid for other properties at the same concentration. Thus, for solutions of electrolytes, P1 = iP1o X1 Pv = in2RT P1 = iP1o X1 ΔTb = iKbM ΔTf =iKfm Application of colligative properties  Preparation of isotonic intravenous and isotonic lacrimal solutions (physiologic saline, for application to the eye to moisten the conjunctiva and cornea).  Determination of the molecular weights of solutes; It used in determining molecular weight is based on the fact that each colligative property is altered by a constant value when a definite number of molecules of solute is added to a solvent. Or, 𝒏𝟐 Since the mole fraction of the solute, 𝑿𝟐 = 𝒏 𝟏 +𝒏𝟐 Where 𝒏𝟐 is the number of moles of solute and 𝒏𝟏 is the number of moles of solvent, 𝒏𝟐 and 𝒏𝟏 are given by 𝑾𝟐 𝑾𝟏 𝒏𝟐 = 𝐚𝐧𝐝 𝒏𝟏 = 𝑴𝟐 𝑴𝟏 𝑾𝟐 is the weight of the solute of molar mass 𝑴𝟐 and 𝑾𝟏 is the weight of the solvent of molar mass 𝑴𝟏. 𝑚𝑎𝑠𝑠 i.e, from the formula; 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  19 𝑷𝒐 −𝑷 Therefore, recall the equation (refer to as the vapor pressure lowering, ΔP) for 𝑷𝒐 example, 𝑷𝒐 − 𝑷 𝒏𝟐 𝑾𝟐 /𝑴𝟐 = = 𝑷𝒐 𝒏𝟏 + 𝒏𝟐 𝑾𝟏 /𝑴𝟏 + 𝑾𝟐 /𝑴𝟐 Since the solution under consideration is very dilute, 𝑾𝟏 𝒏𝟐 < 𝒏𝟏 𝐚𝐧𝐝 𝒏𝟏 + 𝒏𝟐 ≈ 𝒏𝟏 i.e, 𝒏𝟏 = 𝑴𝟏 Thus 𝑷𝒐 − 𝑷 𝑾𝟐 𝑴𝟏 = 𝑷𝒐 𝑾𝟏 𝑴𝟐 𝑷𝒐 𝑾𝟐 𝑴𝟐 = ( 𝒐 ) ( ) 𝑴𝟏 𝑷 − 𝑷 𝑾𝟏 If all the quantities on the right hand side are known, 𝑴𝟐 the molar mass of the solute can be easily calculated. ELECTROLYTIC CONDUCTION Electricity flows through a conductor due to transfer of electrons from a point of higher negative potential to one of lower. However, the mechanism by which this transfer takes place is different for all conductors. There are, first, the metallic conductors or electronic conductors in which the conduction takes place by direct migration of electrons through the conductor under the influence of an applied potential. Metals are typical examples of such conductors. Certain solid salts, e.g., lead sulphide, cadmium sulphide, copper sulphide also belong to this category. Conducting materials of the second type are known as electrolytic conductors or electrolytes. In this type, electron transfer takes place by migration of ions, both positive and negative, towards the electrodes on applying an electrical potential. In other words, various chemical compounds notably acids, bases and salts conduct electricity by virtue of ions present or formed rather than by electrons. An electrolyte therefore is a substance (an acid, a base, or a salt), that in an aqueous solution ionizes to positive ions (cations) and negative ions (anions). A salt such as sodium chloride (NaCl) is an example of an electrolyte. In an aqueous solution, the ions formed are Na+ (a cation) and Cl- (an anion). Therefore, for each molecule, two independent entities are formed PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  20 in solution. In an aqueous solution, electrolytes exhibit the following important properties that differentiate them from non-electrolytes:  They exhibit anomalous colligative properties compared with non-electrolytes.  They can conduct an electric current.  They tend to show rapid chemical reactions compared with non-electrolyte solutions. Electrolytes can be classified into two groups – strong and weak – depending on their ability to ionize in an aqueous solution. Strong electrolytes are substances that are completely ionized in water. Hydrochloric acid, for example, is a strong electrolyte, and its degree of ionization is pH-independent. The weak electrolytes, in contrast, ionize only partly in water. Atropine, Phenobarbital, and sulfadiazine are examples of weak electrolytes. Non-electrolytes are substances that do not ionize in water at all and therefore do not conduct an electric current in solution. Examples of non-electrolytes include sucrose, fructose, urea, and glycerol. Importance of Ionization in Pharmacy Many drugs and pharmaceutical ingredients are organic weak electrolytes. Their degree of ionization is an important physicochemical property and has extensive physiologic implications. Depending on the pH of the environment in which these organic weak electrolytes are present, they can exist either in an ionized form or in an unionized form. This degree of ionization can affect the absorption, transport, and excretion of drugs. In general, the ionized forms are more water-soluble but much less soluble in lipid, while the unionized forms have greater lipid solubility and are more permeable to lipid membranes. The effects of the degree of ionization of organic weak electrolytes on drug absorption lead to the development of the important pH partition hypothesis. According to this hypothesis, the absorption of a weak electrolyte is determined mainly by the extent to which the drug exists in its unionized form at the site of absorption. As an example, the gastrointestinal-blood membrane is impermeable to the ionized form of a drug because of the poor lipid solubility of the drug in this form. A small change in pH can cause a large change in the ratio of ionized to unionized molecules and a correspondingly large change in properties such as solubility, dissolution rate, and partition coefficient. Hence, the degree of ionization can have a profound effect on the absorption, distribution, and elimination of drugs and also can affect factors such as the antimicrobial effectiveness of preservatives, chromatographic separation, and the physicochemical stability of a drug product. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  21 Electrolytes therefore are conductors of electricity; this is due to the ions they contain. The greater the concentration of ions, the higher will be the conductance. Assignment Read about the following  Conductance  Specific resistance  Specific conductance  Equivalent and Molar Conductance  Molecular Conductance  Application of Conductance Measurements  The Conductance Ration  Theories of Strong Electrolytes BUFFER SOLUTIONS These are solutions of compounds or mixtures of compounds which resist changes in their pH upon addition of small quantities of an acid or alkali. Most buffer solutions usually consist of a mixture of a weak acid and one of its salts or a weak base and one of its salts. Buffer Action The ability of certain solutions to resist change in their pH upon addition of an acid or a base is known as the Buffer action. Take an example of a solution of sodium chloride in water. Its pH value is 7. Addition of even 1ml of 1N HCl solution to 1 litre of NaCl solution lowers its pH value from 7 to about 3. Similarly, the addition of even 1 ml of 1N NaOH solution to 1 litre of NaCl solution raises is pH to about 11. Sodium chloride solution is therefore not a buffer. Now let us take an example of a mixture of a weak acid (acetic acid) and its salt (sodium acetate). Acetic acid is very slightly dissociated in solution while sodium acetate being a salt is almost completely dissociated. The mixture thus consists of CH3COOH molecules as well as CH3COO- and Na+ ions. If a strong acid is added to the mixture, the H+ ions supplied by the acid are immediately taken up by CH3COO- ions to form the very slightly dissociated CH3COOH. 𝐇 + + 𝐂𝐇𝟑 𝐂𝐎𝐎 → 𝐂𝐇𝟑 𝐂𝐎𝐎𝐇 PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  22 The H+ ions are thus neutralized by the acetate ions present in the mixture and there is very little change in the pH value of the mixture. On the other hand, if a strong base is added, the OH- ions supplied by the base are neutralized by acetic acid present in the mixture and again there is very little change in the pH of the solution. 𝐎𝐇 − + 𝐂𝐇𝟑 𝐂𝐎𝐎𝐇 ⟶ 𝐂𝐇𝟑 𝐂𝐎𝐎− + 𝐇𝟐 𝐎 A mixture of a weak base and its salt also behaves in a similar manner. Let us take an example of a mixture containing equimolar solutions of ammonium hydroxide and its largely dissociated salt, ammonium chloride. The mixture contains undissociated NH4OH as well as NH4+ and Cl- ions. If a strong acid is added to this mixture, the H+ ions supplied by the acid are neutralized by the base NH4OH. 𝐇 + + 𝐍𝐇𝟒 𝐎𝐇 ⟶ 𝐇𝟐 𝐎 + 𝐍𝐇𝟒+ On the other hand, if a strong base is added, the OH- ions are neutralized by NH4+ ions forming very slightly dissociated NH4OH. 𝐎𝐇 − + 𝐍𝐇𝟒+ ⟶ 𝐍𝐇𝟒 𝐎𝐇 In both the cases there is very little change in the pH. Buffering system of blood Maintain a constant blood pH is critical for the proper functioning of our body. The buffer that maintains the pH of human blood involves a carbonic acid (H2CO3) – bicarbonate ion (HCO3- ) system. This is essential to maintain blood pH between 7.35 and 7.45, a value higher than 7.8 or lower than 6.8 can lead to death. When any acidic substance enters the blood stream, the bicarbonate ions neutralize the hydronium ions forming carbonic acid and water. HCO3- + H3O+  H2CO3 + H2O On the other hand, when a basic substance enters the bloodstream, carbonic acid reacts with the hydroxide ions producing bicarbonate ions and water. H2CO3 + OH  HCO3- + H2O PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  23 In this way, the pH of blood is prevented from becoming either acidic or basic. pH of pharmaceutical formulations The British Pharmacopoeia attaches considerable importance to the pH values of some pharmaceutical preparations like injections, infusions, eye drops etc. they are often required to be at some definite pH value or lie within certain limits of pH after preparation. Hence, it is often necessary to stabilize such preparations using buffers. The following are some of the reasons for such specifications: 1. To increase the stability of the preparation 2. To minimize pain, irritation and necrosis on injection 3. To provide unsatisfactory conditions for growth of micro-organism 4. To enhance physiological activity Buffer Equation The pH of a buffer solution and the change in the pH upon addition of an acid or a base can be calculated based on the use of the buffer equation which is an expression developed by considering the effect of a salt on the ionization of a weak acid or a weak base when the salt and the acid or base have an ion in common. Buffer Equation for a Weak Acid and Its Salt/Common Ion Effect Let us take an example of the effect of addition of sodium acetate on the ionization of acetic acid. Both sodium acetate and acetic acid have an ion common between them, i.e. CH3COO-, the dissociation constant for the acid is given by; 𝐂𝐇𝟑 𝐂𝐎𝐎𝐇 ⟶ 𝐂𝐇𝟑 𝐂𝐎𝐎− + 𝐇𝟑 𝐎+ [𝐂𝐇𝟑 𝐂𝐎𝐎− ][𝐇𝟑 𝐎+ ] 𝐊𝐚 = [𝐂𝐇𝟑 𝐂𝐎𝐎𝐇] The pH of the buffer solution can be obtained by re-arranging the above equation for dissociation constant: [𝐂𝐇𝟑 𝐂𝐎𝐎𝐇] [𝐇𝟑 𝐎+ ] = 𝐊 𝐚 [𝐂𝐇𝟑 𝐂𝐎𝐎− ] PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  24 Since acetic acid ionizes only slightly, the concentration of acetic acid may be considered to represent the total concentration of acid in the solution. Hence, the term [CH3COOH] can be replaced by [acid]. Similarly, since the acetate ion is contributed almost entirely by the salt, sodium acetate, the term [CH3COO-] may be replaced by [salt]. Therefore, [𝐚𝐜𝐢𝐝] [𝐇𝟑 𝐎+ ] = 𝐊 𝐚 [𝐬𝐚𝐥𝐭] Expressing in logarithmic form, 𝐥𝐨𝐠[𝐇𝟑 𝐎+ ] = 𝐥𝐨𝐠 𝐊 𝐚 + 𝐥𝐨𝐠[𝐚𝐜𝐢𝐝] − 𝐥𝐨𝐠[𝐬𝐚𝐥𝐭] Reversing the sign −𝐥𝐨𝐠[𝐇𝟑 𝐎+ ] = − 𝐥𝐨𝐠 𝐊 𝐚 − 𝐥𝐨𝐠[𝐚𝐜𝐢𝐝] + 𝐥𝐨𝐠[𝐬𝐚𝐥𝐭] -log [H3O+] is represented as pH, and -log Ka is represented as pKa. pH = pKa + log [salt] – log [acid] Therefore; [𝐬𝐚𝐥𝐭] 𝐩𝐇 = 𝐩𝐊 𝐚 + 𝐥𝐨𝐠 [𝐚𝐜𝐢𝐝] This is known as the buffer equation or the Henderson-Hasselbalch equation for a weak acid and its salt. The buffer equation is a useful expression used in the preparation of buffered pharmaceutical solutions. It is satisfactory for calculations within the pH range of 4 to 10. In actual conditions, there is a small difference in the pH of the solution since the activity coefficient of the components varies with concentration. Taking into consideration the activity coefficient, the Henderson-Hasselbalch equation can be modified as: [𝐬𝐚𝐥𝐭] or 𝐩𝐇 = 𝐩𝐊 𝐚 + 𝐥𝐨𝐠 [𝐚𝐜𝐢𝐝] + 𝐥𝐨𝐠𝛄𝐀𝐂 Where 𝜸𝑨𝑪 indicates the activity coefficient of the common ion. Example 1 PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  25 Calculate the pH of the buffer solution consisting of 0.1M each of acetic acid and sodium acetate (pka of acetic acid = 4.76). Solution Concentration of acetic acid [acid] = 0.1M Concentration of sodium acetate [salt] = 0.1M pka of acetic acid = 4.76 According to Henderson-Hasselbalch equation, [𝐬𝐚𝐥𝐭] 𝐩𝐇 = 𝐩𝐤 𝐚 + 𝐥𝐨𝐠 [𝐚𝐜𝐢𝐝] 𝟎. 𝟏 𝐩𝐇 = 𝟒. 𝟕𝟔 + 𝐥𝐨𝐠 𝟎. 𝟏 pH = 4.76 Hence, pH of the buffer solution = 4.76. Example 2 What is the pH of a buffer solution prepared with 0.05M sodium borate and 0.005M boric acid? The pKa value of boric acid is 9.24 at 25 oC [𝐬𝐚𝐥𝐭] pH = pka + log [𝐚𝐜𝐢𝐝] [𝟎.𝟎𝟓] = 9.24 + log [𝟎.𝟎𝟓] = 9.24 + log 10 = 9.24 + 1 = 10.24 Example 3 What molal ratio of salt/acid is required to prepare sodium acetate –acetic acid buffer solution with pH of 5.76? The pka value of acetic acid is 4.96 at 25 oC PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  26 [𝐬𝐚𝐥𝐭] pH = pka + log [𝐚𝐜𝐢𝐝] [𝐬𝐚𝐥𝐭] pH - pka = log [𝐚𝐜𝐢𝐝] [𝐬𝐚𝐥𝐭] log = pH – pka [𝐚𝐜𝐢𝐝] [𝐬𝐚𝐥𝐭] log = 5.76 – 4.76 = 1 [𝐚𝐜𝐢𝐝] [𝐬𝐚𝐥𝐭] log [𝐚𝐜𝐢𝐝] =1 [𝐬𝐚𝐥𝐭] [𝐚𝐜𝐢𝐝] = antilog 1 [𝐬𝐚𝐥𝐭] [𝐚𝐜𝐢𝐝] = 10 [𝟏𝟎] = = 10:1 1 PHASE EQUILIBRIA INTRODUCTION PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  27 The phase rule is a generalization which seems to explain the equilibrium existing between heterogonous systems and was theoretically deduced by Willard Gibbs in 1876, it may be stated as; F=C–P+2 Where; F = the degrees of freedom C = the number of components P = the number of phases DEFINATION OF TERMS INVOLVE IN THE PHASE RULE Phase (P) A phase is a homogenous portion of a system with uniform, physical and chemical characteristics separable from the rest of the system by definite boundaries. It is a form of matter that is homogeneous in chemical composition and physical state. Typical phases are solid, liquid and gas. For example, ice, liquid water and water vapor are three separate phases, each is physically distinct and there are definite boundaries between them. Pure liquids or homogeneous solutions (E.g. salt water) are considered as only one phases but two immiscible liquids (or liquid mixed with different compositions) separated by a distinct boundary are counted as two different phases. A mixture of gases always constitutes one phase because the mixture is homogeneous and there is no bounding surface between the different gases in the mixtures Number of components (C) This is the number of chemically independent constituents in a system. It is the number of independent species necessary to define the composition of all phases present in the system. For example, in the three phase system ice, water and water vapor, the number of components is one, because each phase can be expressed in terms of H2O. A mixture of salt and water is a two component system since both chemical species are independent. Degrees of freedom (F) PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  28 The number of degrees of freedom (F) in this context is the number of intensive variables which are independent of each other such as temperature, pressure and concentration/composition that must be specified to define the system completely. Consider any system having one phase only, say gaseous phase of water. In order to define completely the state of the system, it is necessary to specify both temperature and pressure. The system, therefore, has two degrees of freedom or it is said to be bivariant. When two phases are in equilibrium say, ice and liquid water, only temperature or pressure need to be arbitrarily fixed to define the system completely. This system has one degree of freedom or it is univariant. If a system has three phases in equilibrium, such as in the water system in which ice, liquid water and water vapor are together present at triple point, then there are no degrees of freedom because three phases can co-exist in equilibrium only at a particular temperature and pressure which are thus automatically fixed. This type of system is said to be invariant. Limitation of phase rule: - The phase rule gives us very limited information. It only gives the number of degrees of freedom, but does not tell us by how much and in what direction the system moves to achieve the new set of equilibrium. Suppose, we have any system e.g., ice, liquid and vapor in equilibrium, if we change the temperature, the pressure will also change in order to have new set of equilibrium. But phase rule does not tell us by how much, the pressure will change and in what direction the system would be moving. APPLICATIONS OF THE PHASE RULE SYSTEM OF ONE COMPONENT In one component system, C = 1. Therefore, from phase Rule, F=1–P+2 F=3–P … (1) Since the minimum number of phases which can be possible for any system is one, therefore, it follows from equation (1) that the maximum value of degree of freedom for a one component system is two as given by F=3–1=2 So, in order to define one component system, two variables must be specified. The two variables chosen are temperature and pressure. For a one component system, the concentration variable is not taken into account because the composition is always same. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  29 If two phases coexist in equilibrium with each other, then from equation (1), F =1 i.e. the system is monovariant/ univariant and one of the two variables is sufficient to define the system completely. On the other hand, when three phases are in equilibrium with each other, then F = 0, i.e., it is not necessary to specify any variable and the system is non-variant (or invariant). Thus the maximum number of phases which can coexist in equilibrium in one component system is three. These facts, predicted by the phase rule, can be represented graphically by considering temperature-pressure diagram referred to as the phase diagram. A phase diagram is a type of chart used to show conditions at which thermodynamically distinct phases can occur at equilibrium Substances can exist in more than one phase, example, solid, liquid and gas. Some substances even have more than one solid phase, for example Carbon exist as diamond, graphite and mainly other forms related to C60, Tin exist as white Tin and gray Tin etc. A great amount of information on the phases of a substance can be summarized on a phase diagram. Water System Water can exist in three possible forms, namely ice (solid), water (liquid) and water vapor. These three forms can lead to the following equilibria: Two-phase equilibria Solid-liquid Solid-vapor Liquid-vapor Three phase equilibria Solid-liquid-vapor When phase rule is applied to these possible equilibria, three bivariant single-phase areas, three monovariant two-phase equilibrium lines, and one invariant three-phase equilibrium point are anticipated. The phase diagram for the water system is as shown. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  30 Phase diagram for the water The diagram consists of three curves OA, OB and OC representing the equilibrium conditions between two phases and expressing the fact that along the curves, the system is monovariant. These curves divide the diagram into three areas – AOB, BOC and AOC which are respectively the fields of the existence of vapor, liquid and solid. Within the single phase areas, the system is bivariant. O represents the point where the three curves are meeting. At this point, three phases coexist in equilibrium and system is invariant. The study of the phase diagram involves the discussion of the curves, areas and points. This is done as follows: The Study of the Curves Curve OA: - This curve separates the liquid region from vapor region and is known as the vapor pressure curve of liquid water. Along the vapor pressure curve, vapor and liquid coexist in equilibrium. This curve starts from O (the ripple point) and extends up to critical temperature (t1) (the temperature above which the liquid does not exists). This critical temperature is 374 o C and the pressure corresponding to this temperature is 220 atm. This is the temperature above which it is impossible to liquefy water vapor. It can be clearly seen that corresponding to any PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  31 particular temperature, there is only one pressure and vice versa at which the two phases are in equilibrium. The degree of freedom is one and the system is univariant as predicted by the phase rule. F=C–P+2 F=1–2+2=1 Curve OB: - This curve separates the solid region from the liquid region and is called the melting point curve of ice because it tells how the freezing temperature of water or melting temperature of ice varies with pressure in such a way that the two phases are in equilibrium. This curve again starts from O and extends up to very high values of pressure. The degree of freedom of this curve is also one because there is only one pressure corresponding to any temperature at which solid and liquid phases will coexist in equilibrium and vice versa. As can be seen from the diagram, this curve slopes towards the pressure axis. This indicates that the melting point of ice decrease with increasing external pressure. Curve OC: - This part separates the solid region from the vapor region and is known as the sublimation curve of ice, because it gives the vapor pressure of solid ice in equilibrium with water vapor at different temperatures. The curve starts from O, the freezing point of water and extends up to absolute zero. As can be seen from the diagram, there is only one temperature corresponding to any pressure and vice versa at which the two phases are in equilibrium. Hence, the degree of freedom is one. A mass of ice may be converted directly into vapor by heating provided that the pressure is kept below the triple point pressure. This transition is valuable in drying compounds that are sensitive to higher temperatures as in freeze drying. The Areas: - The diagram is divided into three areas, AOC, AOB and BOC, bounded by the lines and shows the regions of ice, vapor and liquid water respectively. Within these single phase areas, the system is bivariant. This means that in order to define completely a point within any of the areas, it is necessary to specify both temperature and pressure. This also follows from the phase rule as shown below: F=C–P+2 =1–1+2=2 The Point O: - We see from the diagram that the three curves OA, OB and OC are intersecting each other at only point O. this is known as the triple point and is obviously the point at which the three phases ice, water, and vapor co-exist in equilibrium. The triple point of water PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  32 corresponds to the minimum pressure at which liquid water can exist. The temperature and pressure corresponding to this equilibrium are 0.0098 oC and 4.58 mmHg respectively. It should be noted that three phases can co-exist in equilibrium only under one set of conditions and neither temperature nor pressure can be changed. If any of the two variables is altered even slightly, the equilibrium is shifted and three phases will not exist together. Hence, the system corresponding to this point is invariant and there is no degree of freedom. This is in accordance with the phase rule. DEGREES OF NUMBER OF SYSTEM PHASES FREEDOM COMMENTS System is bivariant and lies any were marked solid liquid, or vapor. We must fix two variables e.g. P or T to define the Gas, Liquid or Solid 1 2 system. System is monovariant and lies any were along a line between two phase regions, i.e. line AO, Gas–Liquid BO, CO. we must fix Liquid-Solid on variable, either P or T to define the Solid-Gas 2 1 system. System is invariant and lie only at the point of intersection of the lines bounding the three regions, i.e. Gas-Liquid-Solid 3 0 point O SYSTEMS OF TWO COMPONENTS For a two-component system, C = 2 and the phase rule equation can be written as F=C–P+2 F=2–P+2=4–P PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  33 Since the minimum number of phases for any system is one, therefore from equation (1), it follows that the maximum number of degrees of freedom in a two-component system is F=4–1=3 This means that in order to define a two-component system completely, all the three variables- pressure, temperature and concentration must be specified. To represent these relations graphically, we require a three dimensional diagram in which three co-ordinate axes representing pressure, temperature and concentration are at right angles to each other. This will lead to a space model which is difficult to construct on paper. In order to remove this difficulty, the common practice is to fix (or ignore) one of the three independent variables. The various equilibria which are possible for a two-component system are:  Solid-liquid equilibria  Solid-gas equilibria  liquid-liquid equilibria  Liquid-gas equilibria We shall be limiting ourselves to the study of only liquid-liquid and solid-liquid equilibria having no gas phase. Such systems having no gas phase (in which the vapor phase is ignored) are known as condensed systems. In other words, systems in which the vapor phase is ignored and only solid and/or liquid phases are considered are termed condensed systems. Since the effect of pressure is small on this type of equilibria, therefore, experiments are usually conducted under atmospheric pressure, thus keeping the pressure constant. This will reduce the degree of freedom of the system by one, and for such system, the phase-rule becomes F=C–P+1 This is known as the reduced phase rule having two variables, namely temperature and concentration of one of the constituent.  Two component systems containing liquid phases (liquid-liquid equilibria) They consist of partially miscible liquids. They may be divided into the following for convenience of discussions.  System showing an increase in miscibility with rise in temperature e.g. the phenol-water system.  System showing a decrease in miscibility with rise in temperature e.g. the triethlylamine plus water.  System showing upper and lower critical solution temperature e.g. nicotine water system PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  34 System showing an increase in miscibility with raise in temperature Temperature composition diagram for the phenol – water system One of such system is the phenol and water system. The temperature-composition diagram of phenol and water at constant pressure as shown above is convenient to use in the explanation of the effects of partial miscibility in system that shows an increase in miscibility with increase in temperature. The curve ‘gbhci’ shows the limit of temperature and concentration within which two liquid phases exist in equilibrium. The region outside this curve contains systems having only one liquid phase. At point “a” equivalent to a system containing 100% water at 50 oC, adding known increments of phenol to a fixed weight of water at 50 oC will result in the formation of a simple phase until the point “b” is reached, at which point a minute amount of a second phase appears. The concentration of phenol and water at which this occurs is 11% by weight of phenol in water, analysis of the second phase which separates out on the bottom, shows it to contain 63% by weight of phenol in water. This phenol-rich phase is denoted by point C on the phase diagram. As we prepare mixtures containing increasing quantities of phenol, i.e. as we proceed across the diagram from points “b” to point “c” we form systems in which the amount of phenol rich phase (B) continually increases as denoted by the test tubes drawn in the figure above. While the amount of the phenol-rich phase (B) is increasing the water-rich phase (A) is decreasing, once the total concentration of phenol exceeds 63% at 50 oC a single phenol rich phase is formed. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  35 Note that between “a” and “b” there is only one liquid phase. Application of the phase rule shows that there are 3 degrees of freedom F=C – P+2 F=2 – 1+2 =3 This means that temperature and composition most be specified in order to define the system completely at constant pressure i.e (pressure is fixed, so F is reduced to two (2). The maximum temperature at which the two-phase region exists is termed the critical solution temperature (or upper consulate temperature). Above “h” only one liquid phase can exist. The two components are therefore miscible in all proportions above “h” which is known as the upper critical solution temperature. The upper critical solution temperature for phenol water system is 66.8oC, all combination of phenol and water above this temperature are completely miscible and yield one phase liquid system. At a total composition corresponding to point “d” (containing 24% of phenol and 76% of water). We have two liquid phases present in the tube. The water rich phase (A) has a composition of 11% phenol in water while the phenol rich phase (phase B) contains 63% phenol. Phase B will lie below phase A because it is rich in phenol and phenol has a higher density than water. In terms of the relative weight of the two phases, there will be more of the water-rich phase A than the phenol-rich phase B at point “d”. The relative weight of these solutions from the graph is given by Weight of phase A = length dc Weight of phase B length bd The application of the phase rule to the system at point d shows that there are two degrees of freedom; F= 2 – 2 + 2 =2 Again pressure is fix we need only define temperature to completely define the system because F is reduced to 1. From the figure above it is seen that if temperature is given, the compositions of the two phases are fixed by the points e.g. point “b” and “c” at 50 oC. System showing a decrease in miscibility with increase in temperature Example: triethylamine-water system These two liquids are completely miscible in all proportions below 18.5oC, but above this temperature it is partially miscible. The temperature 18.5oC at which both liquids are completely miscible is known as minimum or lower critical solution temperature, because it is the lowest temperature below which the two liquids are completely miscible. Above this temperature all the systems have two liquid layers PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  36 Triethylamine-water system System showing upper and lower critical solution temperature e.g. nicotine-water system. At room temperature nicotine and water are completely miscible in all proportions, but as the temperature is raised above 60.8oC the mutual solubility decreases. If the temperature is raise further as high as 208 oC, the solubility’s once again increases, the two liquid again become miscible. The solubility’s curve would therefore be a closed curve. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  37 Nicotine-water system showing upper and lower consulate temperature The nicotine – water system exhibits two critical solution temperature one upper and another lower. All points outside this curve represents one layer i.e the two layers are completely miscible, while inside, they represent two layers i.e are only partially miscible. The upper or maximum solution temperature is 208oC while the lower or minimum solution temperature is 60oC. Two component systems containing solid and liquid phases (Solid-liquid equilibria) Certain substances such menthol, thymol, camphor, phenol, salol, etc. when mixed in a particular proportion tend to liquefy due to reduction in their respective melting points. Mixtures of such substance are known as eutectic mixtures. (Greek meaning: Eutectic – Easy melting). The two components are completely miscible in the liquid state (when the liquefy) but completely immiscible as solid. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  38 Principle Let us consider two substances, A and B in the figure above. The points A and B represent the melting points of the two components. As increasing quantities of B are added to A, the freezing point of A falls along the curve AC. Similarly, as increasing quantities of A are added to B, the freezing point of B falls along the curve BC. At a particular composition, “C”, known as the eutectic Point, the mixture of the two substances has the lowest melting point. This composition of the two substances is known as the eutectic mixtures. At the eutectic point, three phases (liquid, solid A, and solid B) coexist. The eutectic point therefore denotes an invariant system because, in a condensed system, F = 2 - 3 + 1 = 0. Below the eutectic temperature the mixture of the two substances will exist as solid while above it, the mixture will convert into a liquid. In pharmaceutical practice, eutectic mixtures are difficult to dispense in the form of a powder. In order to incorporate such materials in a powder, it is essential to first mix each ingredient separately with an inert diluent such as light magnesium oxide, magnesium carbonate, starch, kaolin, etc., followed by gentle blending of the different portions. Alternatively, the eutectic materials can first be triturated together in order to force them to liquefy. The liquid can then be adsorbed on inert diluents. The phenomenon of eutectic formation has also been used in pharmaceutical practice to improve the dissolution behaviour of certain drugs. For example, eutectic mixture of aspirin- acetaminophen (37% and 63% respectively), urea-acetaminophen (46% and 54% respectively) and griseofulvin-succinic acid (55% and 45% respectively) dissolve more rapidly than the drugs. It is the principle by which solid dispersions facilitate the dissolution and eventually the bioavailability of poorly soluble drugs when combined with freely soluble “carriers” such as urea or polyethylene glycol. THREE COMPONENT SYSTEMS PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  39 According to phase rule, the degree of freedom in a system of three components is given by: F=C–P+2 F=3–P+2=5–P Since the minimum number of phases that can exist in any system is one, the maximum number of freedom is four. Hence in order to completely define such a system, four variables are required namely, temperature, pressure and concentration of two of the three components. This will require a three dimensional diagram for a complete graphical representation of these systems which is not possible. For this reason, it is customary to represent ternary systems at constant temperature and at constant pressure. By fixing the temperature and pressure, the number of degrees of freedom is reduced by two, so that F = 3 – P and the system has, at most, a variance of two. When temperature and pressure are constant, the remaining variables are concentration variables X1, X2 and X3 of the three components. These concentration variables are related to each other by the equation X1 + X2 + X3 = 1 and can be represented on a planner diagram. If we know any two of them, the value of third can be found. Graphical representation of ternary systems Of the various methods for plotting two dimensional equilibrium diagrams for ternary (three) systems, the method of Stocks and Roozeboom is very popular and most generally employed. In this method, the variation is concentration of the three component mixtures at constant temperature and pressure are expressed by means of an equilateral triangle such as that shown in the figure below. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm  40 Graphical representation of ternary system Each apex of the triangle presents pure component i.e., 100 percent of the component with which it is designated. Each side is divided into 10 equal parts and the lines parallel to BC, CA and AB are drawn. Each of the lines parallel to BC, CA and AB represent various percentages of A, B and C respectively. Any point on any side of the triangle refers to a binary system and two-component gives various proportions of constituents in a two-component system while any point within the triangle represents a ternary system and gives various proportions of any mixture composed of A, B and C. Thus a point ‘x’ on the line AB represents a two-component system of 20 percent A and 80 percent B. similarly; a point ‘Y’ on the line AC represents a two-component system of 40 percent A and 60 percent C. Any point ‘z’ within the triangle represents a ternary system of 20 percent C, 20 percent B and 60 percent A. PCT 201 lecture notes Compiled by Z. S. Yahaya, Ph.D., FPCPharm