Transcript T =
Measurements of dissolution profiles (solubility): • Finely divided powder of known , uniform particle size in excess is suspended in water and stirred continuously. • Samples are taken at time intervals and filtered to remove any undissolved drug, then analysed to determine concentration of dissolved drug. • Plot Ct against time of sampling • When all of the drug is dissolved i.e., solution is saturated with drug the drug concentration will not change as a function of time any further. Measurements of dissolution profiles (solubility): Cs Concentration mg/ml Time (hr) Solubility Product: For slightly soluble electrolytes (eg AgCl & Al(OH)3) dissolved in water to form saturated solutions, their solubility is referred to as solubility product (Ksp). Formation of saturated aqueous solution is an equilibrium process: K AgCl (solid) Ag+ + Cl- [Ag+] [Cl-] K= [AgCl] solid [AgCl] solid is considered constant because there is always some solid present. K = [Ag+] [Cl-] Ksp = [Ag+] [Cl-] Ksp is solubility product Example: A saturated solution of AgCl in water at 20oC has a concentration of 1.12X10-5 M of AgCl ie., Cs of AgCl = 1.12X10-5 M. Calculate the solubility product Ksp. Ksp = [Ag+] [Cl-] Ksp = (1.12X10-5 ) (1.12X10-5 ) = 1.25 X 10-10 For salts that the cations and anions have different number of charges e.g., Al(OH)3 you have to raise each concentration to a power corresponding to the stoichiometric number of ion. K Al(OH)3 (solid) AL+3 + 3OH- Ksp = [AL+3] [OH-]3 In general, More Ksp means more solubility II. Effect of Salts on solubility of solid in liquid: Common ion effect: a) Salting out: K AgCl (solid) Ag+ + Cl- If an ion common with Ag+ or Cl- is added to the solution of AgCl (eg. NaCl), the equilibrium will be shifted towards the left and more AgCl will precipitate leading to reduced solubility. b) Salting in: When the added common ion makes a complex with the salt whereby the net solubility is increased because solubility of new salt is more than solubility of AgCl. Energetics of Solubility Solubility and heat of solution Solubility as a function of temperature for non electrolytes or weak electrolytes or strong electrolytes in non ideal solution can be calculated according to the equation: ∆H solution ln (C= /C- ) = (T= - T-) X R (T= T-) Where C= is concentration at temperature two C- is concentration at temperature one ∆H solution is heat of solution (Cal/ mol) and defined as heat gained when one mole of solute is dissolved in solvent R is universal gas constant (1.9872 Cal/ mol.deg) T= is temperature two in Kelvin = Co +273 T- is temperature one in Kelvin = Co +273 Non electrolytes: Substances that do not yield ions when dissolved in water and therefore do not conduct an electric current through the solution e.g., sucrose or glycerin in water Strong electrolytes: Substances that yield ions by completely ionizing in water when dissolved in water leading to strong conduction of electric current. e.g., hydrochloric acid and sodium sulfate. Weak electrolytes: Substances that yield ions by partially ionizing in water when dissolved in water leading to weak conduction of electric current. e.g., ephedrine and phenobarbital. Ideal solution: A solution where contribution of different components in a liquid mixture to vapour pressure is according to the molar fraction of the different components. Example: The solubility of urea (Mwt. = 60.6 gm/ mol) in water at 25oC is 1.20 gm/ gm water. The ∆H solution for urea in water at 25oC is 2820 Cal/ mol. What is solubility in gm/ gm and the molal solubility at 5oC. Solubility in gm/ gm at 5oC is unknown T- = 5 + 273 = 278 oK T= = 25 + 273 = 298 oK ∆H solution = 2820 Cal/ mol C= = 1.20 gm/ gm R = 1.9872 Cal/ deg. Mol C- = ln (C= /C- ) = (∆H / R) X (T= - T- ) / (T= T-) lnC=- ln C- = (∆H / R) X (T= - T- ) / (T= T-) lnC=- [(∆H / R) X (T= - T- ) / (T= T-)] = ln C- • Example: The solubility of urea (Mwt. = 60.6 gm/ mol) in water at 25oC is 1.20 gm/ gm water. The ∆H solution for urea in water at 25oC is 2820 Cal/ mol. What is solubility in gm/ gm and the molal solubility at 5oC. lnC=- [(∆H / R) X (T= - T- ) / (T= T-)] = ln CLn1.2-[(2820/1.9872)X(298-278 )/(298x278))] = ln Cln C- = -0.161 C- = 0.851 gm/gm the molal solubility at 5oC is unknown. C- = 0.851 gm/gm =851 gm/kg Number of moles = weight / mol. Weight Number of moles =851 gm / 60.6 gm / mol=14.04 Molal solubility = 14.04 mol/kg Solubility at 5oC = 0.851 gm/ gm Solubility at 25oC = 1.2 gm/ gm Thank you