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III. Effect of Temperature Most salts cause absorption of heat during the solution process e.g.,KNO3 and Na2SO4.5H2O and this reaction is called endothermic reaction i.e., heating increases solubility Some salts are not greatly affected by heating e.g., NaCl and this reaction is called isothermic reaction Some salts e.g., Ca-sulfate, Ca-hydroxide and (CH3COO)2Ca.2H20 exhibit exothermic reaction during solution i.e., heating decreases solubility Certain salts e.g., Na2SO4.10H2O exhibit endothermic reaction till temperature reaches 32.4oC then reaction changes to exothermic due to change in crystal structure to Na2SO4 1 2 IV. Effect of pH Most drugs are weak acids or weak bases Acids ionize in alkaline medium, while bases ionize in acidic medium Ionized drug is in the form of salt, conjugate or charged drug The ionized drug is more soluble in water, while the neutral drug is more soluble in other organic solvents e.g., alcohol, chloroform, acetone. Example: Phenobarbital (acidic drug) Increased pH leads to increased ionization leads to increased water solubility and decreased solubility in other organic solvents. Example: Procaine HCl (basic drug) Increased pH leads to decreased ionization leads to decreased water solubility and increased solubility in other organic solvents 3 The fraction of the unionized or ionized form of the drug of a weak acid or weak base is a function of the pH of the solution (medium) and the dissociation constant (ka) of the acid or base (A) Solubility of a weak acid: pH = Pka + log [salt (ionized) / acid (unionized) This is called Henderson Hasselbach equation [HA]T = unionized + ionized [HA]T = [HA] + [A-] in case of saturated solution [HA] = [HA]ss [HA]T = [HA]ss + [A-] [A-] = [HA]T - [HA]ss pHp = Pka + log ([HA]T - [HA]ss) / [HA]ss) 4 This pH is called precipitation pH and defined as the minimum pH at which a weak acid at a given total concentration [HA]T will remain in solution without precipitation i.e., solution stable. Example: A solution contains 1 gm of sodium Phenobarbital in 100 ml at 25oC. Below what pH will precipitation occur given that solubility is 4.78X10-3 M and pka is 7.48 and mol. Weight is 254 gm/mol. [HA]T is 1gm/100ml convert to molar expression =10gm/1000ml number of mole = weight / mol.weight number of mole = 10gm/ 254 gm/mol = 0.0394 in 1000 ml = 0.0394 M pHp = 7.48 + log [(0.0394 - 4.78X10-3) / (4.78X10-3)] pHp = 7.48 + log 7.24 pHp = 7.48 + 0.86 PHp = 8.34 5 If the pH of the solution drops below 8.34, some Phenobarbital will precipitate. A common cause of such a decrease in pH is absorption of atmospheric CO2. (B) Solubility of a weak base: pH = Pka + log [base (unionized) / salt (ionized)] This is called Henderson Hasselbach equation. [B]T = unionized + ionized [B]T = [B] + [BH+] in case of saturated solution [B] = [B]ss [B]T = [B]ss + [BH+] [BH+] = [B]T - [B]ss pHp = Pka + log ([B]ss / [B]T – [B]ss) pHp is the precipitation pH of a base and defined as the maximum pH at which a weak base at a given total concentration [B]T will remain in solution without precipitation i.e., solution stable. 6 Example: Is 8.66 mg/ml procaine solution stable (i.e., no ppt.) at pH 7.4 given that 1 gm dissolves in 200 ml water and pka = 8.05. [B]T = 8.66 mg/ml pka = 8.05 [B]ss = 1gm/ 200ml = 1000mg/200 ml =5 mg/ ml pHp = 8.05 + log (5/ 8.66 – 5) pHp = 8.05 + log 1.37 pHp = 8.19 This is maximum pH and 7.4 is less than 8.19, therefore solution is stable and no ppt. occurs. 7 Pharmaceutical Solvents: (A) Aqueous solvent is water, which has certain advantages: o o o physiologically compatible low toxicity, pharmacologically inactive and tasteless Lack of selectivity is a disadvantage which leads to extraction of unwanted substances. (B) Non-aqueous solvents such as: Ethanol, most widely used solvent besides water, primary solvent for organic compounds Diluted alcohol is a mixture of ethanol and water in the ratio of 1:1 Glycerin, clear syrupy liquid with sweet taste has preservative qualities and used as stabilizer due to its high viscosity 8 Propylene glycol, miscible with water and alcohol good substitute for glycerol Fixed oils such as almond oil, sesame oil, castor oil and olive oil. Other oils such as ethyl oleate and isopropyl myristate Advantages: Non aqueous solvents are mostly used to overcome solubility and stability problems; they are used for sustaining drug release e.g., oily injections; they are used for masking the taste. Disadvantages: Non aqueous solvents have unpleasant taste They have stability problem especially in case of oils 9 Other Methods to improve aqueous solubility: Co-solvency i.e., use of water miscible solvent with water called cosolvents e.g., ethanol, glycerol, propylene glycol and poly ethylene glycol e.g.,Diazepam injection contains 40% propylene glycol and10% ethanol. Use of detergents or surface active agents such as tweens and spans to reduce surface tension e.g., emulsion and suspension Use of complexation using agents such as cyclodextrins to improve water solubility e.g., ketoprofen cyclodextrin complex Use of prodrugs which are water soluble chemical derivatives synthesized to improve water solubility and rapidly regenerate the parent drug at the site of action. E.g., hydrocortisone phosphate and hydrocortisone succinate. 10 Colligative (also called additive) Properties of Solutions: These are physicochemical properties which arise from the addition of solute to solvent to form a solution as opposed to the pure components and they depend on the relative proportions of the molecules of the solutes and solvents. The colligative properties of solutions are: 1) Lowering of vapor pressure Definition of vapor pressure: tendency of solution to evaporate i.e., change from liquid to gas phase. When a non volatile solute is dissolved in a volatile solvent, the relative vapor pressure lowering is given by Rault’s law as follows: P/po = n2 / (n1+n2) where: P is vapor pressure lowering po is vapor pressure of pure solvent P/po is the relative vapor pressure lowering 11 n1 is number of moles of solvent n2 is number of moles of solute Example: Calculate the relative vapor pressure lowering at 20° C for a solution containing 171.2 gm sucrose in 1000 gm of water. M.wt. of sucrose is 342.3 gm/mole and M.wt. of water is 18.02 gm/mol. Number of moles of sucrose (n2) = 171.2 ÷ 342.3 = 0.5 moles Number of moles of water (n1) = 1000 ÷ 18.02 = 55.5 moles P/po = n2 / (n1+n2) P/po = 0.5 / (55.5 +0.5) = 0.0089 OR 0.89% i.e., vapor pressure has been lowered by 0.89% by the addition of 0.5 mole of sucrose. 12 2) Elevation of boiling point: Normal boiling point is defined as the temperature at which the vapor pressure of the liquid becomes equal to an external pressure of 760 mm Hg (1 atmosphere). When you add a solute to a solvent the boiling point is increased as described by equation: TP = Kb . m Where: TP is boiling point elevation Kb is molal elevation constant (m) is molal concentration Example: Calculate the boiling point elevation when a drug is dissolved in water to give a 0.200 molal aqueous solution given that Kb for water is 0.515 deg.kg/mol. 13 TP = Kb . m TP = 0.515 deg.kg/mol X 0.2 mol/kg = 0.103° C. i.e., 0.2 molal solution increases the boiling point by 0.103° C. 3) Depression of Freezing point: The normal freezing point or melting point of a pure compound is the temperature at which the solid and liquid phases are in equilibrium under a pressure of one atmosphere. Equilibrium means the tendency of the solid to pass into liquid state is the same as tendency of the liquid to pass into solid state (e.g., Tertiary butanol). 14 Freezing point depression can be calculated according to equation: Tf = Kf (1000W2 /W1.M2) Where: Tf is Freezing point depression and Kf is molal depression constant W1 is weight of component one (water) W2 is weight of component two (solute) M2 is Molecular weight of component two (solute) Example: What is the freezing point depression of a solution containing 3.42 gm of sucrose and 500 gm of water. The molecular weight of sucrose is 342 gm/mol and Kf is 1.86. Tf = 1.86 (1000 X 3.42 /500 X 342) Tf = 0.037oC i.e., the freezing point of 500 gm water was decreased by 0.037° C by addition of 3.42 gm of sucrose. 15 4) Osmotic Pressure KMNO4 Diffusion solution membrane Solvent KMNO4 Water molecules water If the diffusion membrane is permeable then color of solvent will turn pink due to diffusion of KMNO4 molecules into solution and diffusion of water molecules into KMNO4 solution. 16 But if we used a semi permeable membrane (a membrane that allows only water molecules to pass through it), then only water molecules pass through membrane a phenomenon called osmosis. Definition of osmosis: Osmosis is defined as the passage of the solvent into a drug solution through a semi permeable membrane. Definition of osmotic pressure: Osmotic pressure is the driving force of osmosis and defined as the tendency of the solvent molecules to pass into a drug solution through a semi permeable membrane. 17 Osmotic pressure can be calculated according to equation: π = (RT/V1) ln (P0/P) Where; π is osmotic pressure in atmosphere R is gas constant = 0.082 lit.atm/mol.deg T is Absolute temperature (oK) V1 is molar volume of water P0 is vapor pressure of water P Is vapor pressure of solution Example: Calculate the osmotic pressure for a 1 m aqueous solution of sucrose given that the vapor pressure of the solution is 31.207 mm Hg and vapor pressure of water is 31.824 mm Hg at 30o C. The molar volume of water is 0.0181 L/mol. Temperature = 30 + 273 = 303o K π = (RT/V1) ln (P0/P) π = (0.082 X 303/ 0.0181) ln (31.824/31.207) π = 27.0 atm 18 Isoosmotic and Isotonic Solutions: Isoosmotic solutions: Two solutions that have same osmotic pressure. E.g., an isoosmotic solution with RBC’s is a solution that has same osmotic pressure as RBC’s which equals 280 mosmol/L Isotonic solution: Solution with same osmotic pressure as RBC’s provided that RBC membrane is true (perfect) semi permeable membrane i.e., allows only passage of H2O molecules and not drug molecules. E.g. 2% boric acid solution has same osmotic pressure as RBC’s , therefore it is isoosmotic with blood, but boric acid molecules can diffuse through RBC membrane (not true membrane for boric acid) i.e., this solution is not isotonic causing rapid heamolysis of RBC. 19 E.g. 2% boric acid eye solution is isoosmotic and isotonic with eye because mucous lining of the eye acts as true membrane i.e., boric acid molecules can not pass i.e., it is isotonic with eye. Example of isotonic solution: 0.9 % NaCl solution has same osmotic pressure as RBC’s and RBC membrane does not diffuse Na Cl molecules (true membrane). Examples of non isotonic solutions: 2% NaCl solution is hypertonic solution causing RBC shrinkage 0.2 % NaCl solution is hypotonic solution causing RBC heamolysis 20 Diffusion: It is defined as mass transfer of individual molecules of substance caused by random molecular motion and associated with concentration gradient. Methods of Diffusion Permeation through non porous media i.e., through bulk membrane. Permeation through pores and channels. Fick’s first law of Diffusion Flux (the amount of material flowing through unit cross-section of a barrier in unit time) is proportional to the concentration gradient according to equation: 21 Fick’s first law of Diffusion J=-D dc dx Where: J is flux D is diffusion coefficient in cm2/sec which is a measure of how fast the drug molecules move or diffuse through the membrane C is concentration X is distance in cm of movement perpendicular to the surface of barrier. Fick’s Second law of Diffusion It examines the rate of change of diffusant concentration at a point in the system and states that change in concentration with time in a particular region is proportional to change in concentration gradient at that point in the system according to equation: 22 Fick’s Second law of Diffusion dc = D dt d2c dx2 Where: D is diffusion coefficient in cm2/sec which is a measure of how fast the drug molecules move or diffuse through the membrane C is concentration T is time X is distance in cm of movement perpendicular to the surface of barrier. 23 Polymorphism: • Solid substances can exist as amorphous or crystalline • Polymorphism describes one solid component that can exist in more than one crystal form and even though polymorphs are chemically identical, each form has different physical properties such as, melting point, solubility and density. • Examples of drugs which exists as polymorphs are: barbiturates, chloramphenicol palmitate, steroids, long chain fatty acids, triglycerides and sulfonamides. • Conversion of one crystal form of a drug to another can significantly affect its behavior and usefulness e.g., one crystal form of chloramphenicol palmitate dissolves in GI fluids very slowly that the drug is poorly absorbed, yet another form of the drug dissolves rapidly and well absorbed orally. 24 B) Solubility of Gases in Liquids: Examples: CO2 in H2O Gas in liquid Air in H2O Gas in liquid O2 in H2O (water saturated oxygen) Liquid in gas Nitrous oxide (N2O) in O2 Gas in gas Factors affecting solubility of gas in liquid: 1) Pressure Effect of pressure is expressed by Henry’s Law which states that: In a very dilute solution at constant temperature, the concentration of the dissolved gas is proportional to the partial pressure of the gas above the solution at equilibrium. 25 Liquid Significance: Solubility of a gas in a liquid increases as the pressure on the gas increases. Some times solubility of a gas is decreased so that the dissolved gas escapes violently when the pressure above the solution is released as in gas tanks, so these tanks should be opened slowly. Similar phenomenon is recognized in effervescent solutions when the stopper of the container is removed. 26 2) Temperature: Increasing temperature of a dissolved gas leads to decreasing solubility of most gases due to greater tendency of the gas to expand at higher temperature. Therefore we should use caution when opening containers of gas at high temperature. e.g., ethyl nitrite vessel should be immersed in cold water or ice before opening. 3) Addition of salts (salting out): Gases dissolved in liquids are often liberated from solution in which they are dissolved in by the introduction of an electrolyte such as NaCl or non electrolyte such as sucrose and this phenomenon is called salting out. The added salt or non electrolyte reacts with water which reduces the density of the aqueous environment adjacent to the gas molecules due to consumption of water. 27 C) Solubility of Liquids in Liquids: They are classified according to number of components as: binary solution (2 components e.g., ethanol and water) or ternary components (3 components e.g., water, glycerol and ethanol). They are also classified according to their solubility (miscibility) as: 1) Completely soluble e.g., alcohol in water, glycerin in water. 2) Partially soluble e.g., phenol in water, ether in water. 3)Totally not soluble e.g., castor oil in water, liquid petrolatum in water. 28 Effect of temperature on solubility of liquid in liquid: Partial vapor pressure is defined as the tendency of the molecules to escape from the liquid phase to the vapor phase. As you increase the temperature of the liquid-liquid solution, more molecules will escape and vapor pressure will increase until it becomes equal to the external pressure above the liquid and this temperature is defined as the boiling point. Then increasing temperature above this point will lead to evaporation. i.e., increasing the temperature reduces solubility of liquid in liquid. Liquid in liquid 29