Dr. Toh Compiled L1-L10 PDF - Physical Pharmacy I

PHYSICAL PHARMACY I Course Code: LECTURER: DR. TOH SEOK MING FAR 221/3 Academic Session: Semester I, 2014-2015 Lecture 1 Introduction Is this something you’ve learnt so far? “Substances and compounds are chemicals which can be classified into organics and inorganics....

PHYSICAL PHARMACY I Course Code: LECTURER: DR. TOH SEOK MING FAR 221/3 Academic Session: Semester I, 2014-2015 Lecture 1 Introduction Is this something you’ve learnt so far? “Substances and compounds are chemicals which can be classified into organics and inorganics.” Yes Maybe No  Is this something you’ve learnt so far? “Whenever you come across a substance or compound, always Polarity: categorize them Very polar according to polarity.” Polar Semipolar Unpolar Very unpolar Is this something you’ve learnt so far? “Substances and compounds interact through bonds.” “The term Van der Waals forces is a general term for those intermolecular forces that include dipole–dipole and London forces.” Van der Waals forces are the weak attractive forces in a large number of substances. Next  “Polar molecules can attract one another through dipole–dipole forces.” The dipole–dipole force is an attractive intermolecular force resulting from the tendency of polar molecules to align themselves such that the positive end of one molecule is near the negative end of another. Also called Keesom Forces Next “London forces are the weak attractive forces between molecules resulting from the small, instantaneous dipoles that occur because of the varying positions of the electrons during their motion about nuclei.” Also called Dispersion or Induced dipole- induced dipole Forces Next Concept check Is this something you’ve learnt so far? “The strength of bonds or interaction and the distance between molecules determine their existence as solid, liquid or gas.” Solid Liquid Gas Strength of Strong Intermediate Weak interaction Distance Closely packed Nearer Far between molecules Movement of Not moving Free Very free molecules (only vibrate)  Is this something you’ve learnt so far? “Solids can be crystalline or amorphous.” Composed of one or more crystals Has a disordered structure; lacks well- Each crystal has a well-defined ordered defined arrangement of basic units (atoms, structure in three dimensions molecules, or ions) found in a crystal Glass is an amorphous solid obtained by Sodium chloride (table salt) and sucrose cooling a liquid rapidly enough that its basic (table sugar) are examples of crystalline units are “frozen” in random positions substances. before they can assume an ordered Metals are usually compact masses of crystalline arrangement. crystals A natural example is obsidian, formed when molten rock from a volcanic eruption cools quickly. Crystalline Amorphous solid solid  Is this something you’ve learnt so far? “When two compounds interact, either new compound(s) or solution is formed.” Form new compound(s) Gas/Gas -chemical bonding Gas/Liquid When 2 compounds interact Form solution -solubility may be high or low depending on solute and solvent polarities Liquid/Liquid Solid/Liquid Liquid/Solid Solid/Solid -may form eutectic mixture Concept check Identify the solute(s) and solvent(s) in the following solutions. a. 80 g of Cr and 5 g of Mo b. 5 g of MgCl2 dissolved in 1000 g of H2O c. 39% N2, 41% Ar, and the rest O2 Concept check Give an example of a solid solution prepared from a liquid and a solid. Solution A dental filling, made up of liquid mercury and solid silver, is a solid solution Is this something you’ve learnt so far? “If solute and solvent are from similar types – dissolution is easy and solubility is high.” Therefore, solubility value of a solute in a solvent is a good indication of their chemical characteristics and interaction Concept check Which of the following compounds is likely to be more soluble in water: C4H9OH or C4H9SH? Explain. Solution C4H9OH, because of hydrogen bonding.  TOH Surface free energy and tension Complexation between chemicals to Partition form bigger or coefficient more complex chemicals TOH Formation of soluble and insoluble Ionization Solubility monolayers at surface or interface TOH Colligative Adsorption onto properties solid TOH Diffusion  Diffusion Is a process where molecules, atoms or ions move freely through Brownian motion, resulting in a net effect of their migration from an area of higher concentration to an area of lower concentration which continues until the concentration at all areas are the same, after which the movement still continues but the magnitude to all areas are the same. Diffusion is also defined as a process of mass transfer of the individual molecule of a substance brought about by the random molecular motion associated with a driving force such as the concentration gradient.  Example 1: Water Blue solution Solid CuSO4 (blue)  Example 2: Across a membrane Coloured solution Water 1st half 2nd half of Membrane diffusion cell Example 3: Dark ink Agar Coloured agar Example 4: Liniment containing volatile oil Slimmer After months Plastic bottle Example 5: i) Wet if contains hygroscopic material or deliquesce material ii) Change colour if Tablet in - Hydrolysed by water vapour plastic or bottle - Oxidised by oxygen Methods for Studying Diffusion in Liquid Free boundary method No boundary or separation between solute and solution Diffusion through gel Diffusion through plate of ceramic filter or sintered Involves boundary glass filter Diffusion through membrane  Fick’s First Law The net diffusion rate (dm/dt) is: -Proportional to the difference in concentration between two points (the concentration gradient, dC) -Proportional to the area across which diffusion occurs (A) -And inversely proportional to the distance between two points in the space (dx) Fick’s First Law Diffusion coefficient Mo - Mx dm dC = -DA dt dx Negative value because the solute diffuses from high concentration to area of lower concentration  Cooper and Woodman Equation (1946) 2 M x = Mo e /4Dt) (-x ln Mx = ln Mo -x2/4Dt x2/4Dt = ln Mo – ln Mx x2/t = 2.303 x 4D (log10 Mo – log10 Mx) Fill the top of gel with test solution (Mo). Visually follow migration of dye and match Mo coloured edge with a tube of known x concentration. Measure x (distance traveled which = concentration Mx) at Mx interval t. Lastly plot a graph for x2 vs t.  Cooper and Woodman Equation Should be a straight line with a slope of 2.303 x 4D (log10 Mo – log10 Mx) x2 t Diffusion through a plate or membrane i) Solution Sintered glass filter C = Co Solvent ii) iii) Diffusion cell Solution Solution Solvent C = Co C = Co Solvent Membrane m = -DA(C2 – C1) x (t2-t1) A = area of barrier C2 = concentration at t2 L L = thickness of barrier C1 = concentration at t1 in receiving solution  Other related equations i) Einstein K = Boltzmann’s constant D = KT T = absolute temperature (K) f f = frictional coefficient According to Stokes, r = molecular radius  = viscosity of the medium f = 6r ii) Stokes-Einstein D = KT K=R R = gas constant 6r N N = Avogadro number iii) Sutherland-Einstein D = RT N6r  Diffusion coefficient, D Fick’s First Law Electrokinetic Area, A interaction Pore size Membrane-polymer (concentration and Concentration branching) gradient, dC Plate or disc- ceramic or silica Shape of solute Spherical Diffusion Time, dt Non-spherical Distance Viscosity of traveled, dx; or solvent,  thickness of barrier, L Stokes-Einstein Radius of solute, Temperature, T r Frictional coefficient, f Polymer branching ↑branching of polymer ↓ compactness ↑ pore size ↓ frictional coefficient ↓ interaction of solute with polymer ↑ diffusion  Electrokinetic interaction At low pH At pH = isoelectric point At high pH (slightly positive) (neutral) (negative) Retards diffusion of negatively charged Retards diffusion of positively charged solute solute i) Agar: an acidic polysaccharide Having –O-SO3H group on chain of polymer  Electrokinetic interaction At pH < isoelectric At pH = isoelectric At pH > isoelectric point point point (cation) (neutral) (anion) Will retard diffusion of Will retard diffusion of negatively charged solute positively charged solute ii) Gelatin: an amphoteric polymer 4Hyp Gelatin contains many glycine (almost 1 in 3 Gly residues, arranged every Gly third residue), proline and 4-hydroxyproline Ala Gly Pro residues. A typical Glu Pro structure is -Ala-Gly-Pro- Arg-Gly-Glu-4Hyp-Gly- Pro-. Arg Concept check State true or false The diffusion process in liquids a. is described by Fick's First Law. True b. is faster in higher temperatures. True c. is slower in higher temperatures. False d. does not depend on the pressure True applied. e. depends on the size of the diffusing True molecule. Review questions 1. Which of the following statements is not in accordance with Fick’s First Law? a) The rate of diffusion is directly proportional to the concentration gradient. b) The rate of diffusion is directly proportional to the diffusion coefficient. c) The diffusion coefficient is independent of concentration. d) The rate of diffusion varies with time. 2. The driving force for diffusion to occur is: a) The concentration gradient b) The temperature gradient c) The force gradient d) Diffusion does not require a driving force  Uses of Applications of Diffusion Studies Determine antimicrobiological Determine physical activity of antibiotics parameters of solutes from diffusion in gel experiment Separation and purification of material(s) from a mixture  Determine physical parameters of solutes Stokes-Einstein If know D and  (viscosity) at T of studies, we can calculate r of diffusate Based on: From r, can calculate volume, Important condition: Diffusate or solute must not interact From volume and density (ρ), we can with gel, disc, filter calculate mass, or membrane used in the experiment. Any interaction Molecular weight, MW = m x N leads to wrong results.  Determine MIC of antibiotics from diffusion in gel experiment Assays for bacteriostatic activity such as disc diffusion Cooper (qualitative) and E-tests (quantitative) employ chemically and Woodman defined media (e.g. Mueller-Hinton at a pH of 7.2–7.4, and agar plates of defined thickness. These tests may be used for the determination of MIC. Based on: “Zone of Inhibition” The zone of inhibition is dependent on antibiotic diffusion through agar gels previously seeded with a test organism. After incubation, the organisms fail to grow where the antibiotic has reached a sufficiently high concentration so that the zone of inhibition can be correlated with antibiotic potency.  Molecular weight Diffusion Size or coefficient radius Frictional Shape and coefficient volume Adsorptive Polarity/ capacity at charge at different different physical pH & conditions solvent Partition Solubility coefficient at between solvent & other different phase at pH & different pH solvent  Chromatography Chromatography is a technique for separating and analyzing a mixture of gases, liquids or dissolved substances. This is achieved by using two immiscible substances, one of which (the mobile phase) transports the sample mixture through the other (the stationary phase). The velocity of each material will depend on their affinity to the mobile phase and stationary phase High affinity to Less affinity to High affinity to Less affinity to mobile phase mobile phase stationary phase stationary phase Move faster Move slower Move slower Move faster  Solubility Adsorptive ability Affinity Polarity Partition coefficient  Molecular weight Speed of Diffusion movement Molecule size and coefficient of volume materials Molecule shape and frictional coefficient Thin layer chromatography The apparatus for paper chromatography is shown. A piece of filter paper is the stationary phase and a chosen solvent the mobile phase. Tiny spots of the sample are applied to the paper, which is then developed. The different coloured components of the mixture separate as the solvent rises. The stationary phase may include:  Paper e.g. filter paper  Thin layer of alumina, silica or other materials on a glass, plastic or metal plate  Thin layer chromatography Analysis of drugs (hashish/ cannabis/ marijuana/ ganja) via TLC. The mobile phase moves up the thin layer through capillary effect carrying material having highest affinity to it, followed by those with lesser affinity leaving spots of separated materials along its movement. (Photos by Ulrich Baumgarten via Getty Images)  Basic Column Chromatography 1. Support goes into a column 2. Equilibration 3. Sample binding 4. Column wash 5. Sample elution 6. Column regeneration (+/-)  Liquid Chromatography Fill with mobile phase. Let it flow down and level with the surface of stationary phase. Close tap. Fill with mixture or extract, open tap and let level with surface of stationary phase. Close tap. Fill with mobile phase, open tap and let the mobile phase flow and continue filling the mobile phase until all fractions are collected. Chromatography scale High Performance Liquid Chromatography A nutrition professor at Tufts University measures vitamins and minerals using HPLC (Photo by Wendy Maeda/The Boston Globe via Getty Images)  HPLC A research scientist holds a vial of cancerous blood serum with a graph on the monitor showing a chromatogram of normal serum compared to cancerous serum. (Photo by Gerald Martineau/The Washington Post via Getty Images) How Does High Performance Liquid Chromatography Work? High-Performance Liquid Chromatography [HPLC] System Isocratic LC System High-Pressure-Gradient System The mobile phase composition changes during the separation. This mode is useful for samples that contain compounds that span a wide range of chromatographic polarity. A Typical HPLC [Waters Alliance] System Understanding How a Chromatographic Column Works – Bands How Peaks Are Created Identification Quantitation Dialysis The separation of smaller molecules from larger molecules or of dissolved substances from colloidal particles in a solution by selective diffusion through semi-permeable dialysis membranes or tubes. The dialysis membranes or tubes are made of polymers. Polymers have Different molecular weight Different shape Dense if straight Different packing Loose if branched Different pore size Different charges Different hydrophilicity and hydrophobicity Therefore, one can separate solutes or diffusates according to size, hydrophilicity and lipophilicity  Dialysis membranes are used as filters during reverse osmosis To purify water preventing particles, molecules and ions bigger than pore size from passing through  Dialysis membranes are used as filters during dialysis of untrapped drug Drug can be entrapped in Nanoparticles Nanocapsules Nanoemulsions Liposomes Niosomes Some of drug remained outside of vehicles (untrapped) and therefore must be removed Dialysis membranes are used as filters during hemodialysis (dialysis of blood) To purify the blood plasma of patients suffering from renal failure Hemodialysis (dialysis of blood) Diffusion studies have many uses when it comes to drugs and pharmaceuticals  The experimental setup We can have drug alone or We can use with other various types drugs of membranes to achieve the objective We can also vary the pH of donor and/or receiver solution to mimic site of absorption Diffusion studies for drugs and pharmaceuticals To study absorption Can use biomembrane or synthetic membrane which mimics the biomembrane Use pH similar to intestine if want to predict what happens in intestine Can have other drug if want to see if the absorption is altered by other drug Can add other materials e.g. antacid, kaolin and other insoluble substances to see if insoluble solid affect the absorption of a drug  Diffusion studies for drugs and pharmaceuticals To predict drug release from polymer-encapsulated or - coated dosage form Study the transfer (diffusion) through the membrane Compare with membranes made from other polymers or their mixture Select best polymer or polymer mixture for actual coating based on the diffusion or transfer rate Fast transfer Good if want immediate and short action Not good for extended or slow release Slow transfer Good for extended or slow release Very slow May not be good as drug released very slowly and the concentration in blood may not reach the level for therapeutic effect  Diffusion studies for drugs and pharmaceuticals To study release of drugs from a matrix Sustained release of drugs from tablets has been obtained by incorporating the drug in an insoluble matrix such as plastic resin, wax and fatty alcohol. When brought into contact with the GI fluid, the drug particles slowly leach out as the fluid penetrates the pores of the tablet. Diffusion of the solute through the liquid-filled tortuous paths plays a significant part provided that the tablet retains its framework during the process. Higuchi (1963) suggested that the rate of release of drug from one surface of an insoluble matrix would be governed by the relationship: DCs 1/2 Q=[ (2A – Cs)t]   Higuchi Equation DCs 1/2 Q=[ (2A – Cs)t]  Q : Amount of drug released per unit area at time t D : Diffusion coefficient  : Porosity of the matrix Cs : Solubility of drug A : Concentration of the drug in the tablet  : Tortuosity of matrix Review questions 1. The stationary phase in thin-layer chromatography is: a) liquid held between glass b) silica gel c) glass plate d) none of the above 2. A combination of paper chromatography and electrophoresis involves: a) partition chromatography b) electrical mobility of the ionic species c) both (a) and (b) d) none of these Review questions 3. Predict the elution order of the following solutes in reversed phase HPLC. (Clue: it uses a hydrophobic stationary phase) b>a>c>d 4. Generally, how do you make a calibration curve? a) By using a RI-detector and sucrose as analyte. b) Injecting a known amount of analyte and determine the peak area. c) Accurately measure the flow rate. d) Setting the detector response to zero. Review questions 5. Chromatographic retention is due to: a) Different injection times by the autosampler. b) Adsorption of the analyte to the stationary phase. c) Differences in absorbance in the UV detector. d) Deviations in the flow from the pump. 6. What is characteristic of gradient chromatography? a) The detector scans over a wide range of wavelengths. b) The mobile phase is unbuffered. c) A gradient of flow velocity is created. d) A gradient of mobile phase composition is created. Review questions 7. Describe the process of reverse osmosis that is used to desalinate seawater. Answer: Water is fed, at high pressure, through tubes of semipermeable material. The water passes through the tubing material and the ions do not. 8. A patient has recently undergone renal failure and the only available replacement therapy available is hemodialysis. The dialyzer was arranged in countercurrent flow. What is the advantage of a counter-current flow arrangement over a co-current one? Answer: There is a lower clearance when the hemodialyzer uses a co-current flow arrangement. In co-current flow, solutes are travelling in the same direction as the blood after they have been transferred into the dialysate. This results in a decreased concentration gradient down the length of the dialyzer. In countercurrent flow, the solute is carried away in the opposite direction by the dialysate. Subsequently, a greater average concentration gradient is maintained. This results in a greater clearance as more solute will be removed. Review questions 9. Your friend is trying to convince you that using blood as dialysate fluid would be better than anything. Although he worries that two different blood types could potentially interact. Explain to your friend why there would be no reaction, why blood would be a poor choice and what is actually used (the major components) for dialysate. Answer: If blood was used as a dialysate fluid for hemodialysis there could not be any reactions between blood types because the blood would never actually contact each other. The blood is separated by a thin membrane which allows for some transport but red blood cells which contain the antigens, and antibodies which respond to the antigens do not cross the membrane. Blood would be a poor choice for a dialysate because you require a difference in concentration between the blood and the dialysate. The blood on both sides of the membrane would be almost exactly the same in concentration and there would be little to no flow of the waste materials out of the person on dialysis. The main components of dialysate fluid are pure water, electrolytes, buffer, and sugar.  Section 1 States of Matter or Materials at various Section 2 temperatures (T) and pressure (P) Section 3 Section 4 Section 5 Section 1 Section 2 Section 3 Section 4 Section 5 Phase diagram at various T and P Section 1 Section 2 Section 3 Section 4 Section 5 The Solution Phase Diagram Section 1 Section 2 Section 3 Section 4 Section 5 Triple point (tert-butyl alcohol)  Section 1 At points above Triple Point At fixed P At fixed T Section 2 ↑ of T ↑ of P SolidLiquid GasLiquid LiquidVapour LiquidSolid Section 3 ↓ of T ↓ of P LiquidSolid LiquidGas VapourLiquid SolidLiquid Section 4 Section 5  Section 1 At points below Triple Point Section 2 At fixed P At fixed T ↑ of T ↓ of P Section 3 SolidVapour SolidVapour Section 4 Section 5  Section 1 The problem with traditional drying Section 2 Difficult to remove water completely Thermo labile material - Heat changes the Section 3 shape, texture and composition of material – taste, smell and appearance Section 4 Freeze-drying – locks in original composition and structure without heat (evaporation) – converts solid water (ice) to water vapour, thus skips liquid phase Section 5  Section 1 Lyophilization (freeze-drying) Section 2 Completely removes water from material, i.e. food, biological samples, while leaving basic structure and composition of material intact Section 3 Reasons: Removal of water keeps sample from spoiling for a long time Section 4 Reduces total weight of sample – transport- military (MRE), space Pharmaceuticals Section 5 Research – preserves samples Section 1 Section 2 Section 3 Section 4 Section 5 The Lyophilization Process  Section 1 Effect of Solute on Phase Diagram Section 2 Section 3 At fixed P Section 4 Addition of solute causes Boiling point to increase from Tb1 to Tb2 Therefore, ΔTb = Tb2 – Tb1 (increase in boiling point) Section 5 Freezing point to decrease from Tf1 to Tf2 Therefore, ΔTf = Tf1 – Tf2 (decrease in freezing point)  Section 1 Effect of Solute on Phase Diagram Section 2 Section 3 At fixed T Section 4 Addition of solute causes Freezing pressure to increase from Pf1 to Pf2 Boiling pressure to decrease from Pb1 to Pb2 Reflecting a decrease in ability to produce vapour Section 5 i.e. more difficult to change into vapour. Therefore, vapour pressure decreases  Section 1 Conclusion: Colligative Properties - changes of parameters or Section 2 properties upon the addition of solute Section 3 ↑Tb ↓Tf Section 4 ↓Pb ↑Pf Section 5 ↑ (osmotic pressure)  Colligative properties are the properties of a solution which are affected or changed with the concentration or addition of solute Depending on the quantity of particles (molecules or ions) of solute but not their nature (chemical or physical) Colligative Properties of Solution return to Definition menu Colligative Properties of Solution Boiling Freezing Increase Decrease Point Point in osmotic in vapour Elevation Depression pressure pressure Boiling Point Elevation ΔTb= Kb.i.m – Kb - molal increment constant of solvent boiling point, ebullioscopic constant, boiling-point- elevation constant (for H2O = 0.512 °C/m) – i – Van’t Hoff factor (number of particles into which the solute dissociates) – m – molality (moles of solute per kg of solvent) m↑ → ΔTb↑ → Tb of solution ↑ Uses: i. Addition of solid/ solute → ↑ Tb of volatile oils (peppermint, eucalyptus). Therefore their loss during preparation involving heat can be reduced ii. Adding solutes (e.g. ethylene glycol) in cooling liquid of radiator → ↑ Tb, thus avoiding cooling liquid from loss through boiling or vaporization 1 2 3 4 Defining the Van’t Hoff Factor, i observed colligative property i = calculated colligative property normal MW (assuming no assoc. or dissoc.) i = abnormal MW (experimentally determined) actual number of particles in solution after assoc. or dissoc. i = number of formula units in solution before assoc. or dissoc. The Van 't Hoff factor (named after Jacobus H. Van 't Hoff) is the ratio between the actual concentration of particles produced when the Definition substance is dissolved, and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, i is essentially 1. For most ionic compounds dissolved in water, i is equal to the number of discrete ions in a formula unit of the substance. The Van't Hoff factor does not have to be a whole number. E.g. for substances that do not completely ionize. 1 2 3 4  If association takes place i decreases If dissociation takes place i increases The number of moles of i = | [ 1 + ( y-1 ) x ] | products formed is related to the degree of ionization or association and is given by the Where y = number of products formed Van't Hoff factor. x = degree of ionization or association Example K4[Fe(CN)6] gives rise to 4K+ and the rest. x= 20% Therefore, i = | [ 1+ (y - 1) x ] | = | [ 1 + (5 - 1) 1/5] | = 1.8 1 2 3 4  Freezing Point Depression ΔTf = Kf.i.m – Kf - molal decrement constant of solvent freezing point, cryoscopic constant, freezing-point- depression constant (for H2O = 1.86 °C/m) – i – Van’t Hoff factor – m – molality m↑ → ΔTf↑ → Tf of solution ↓ Therefore, solution difficult to freeze or solidify i.e. will remain as liquid at lower temperature Uses: i. Prevent products e.g. emulsions, lotions etc from solidifying at low temperature ii. Prevent cooling liquid from freezing, protect engine iii. Prevent fuel e.g. diesel from solidifying iv. Preparation of isotonic solution Isotonic solutions have colligative properties equivalent to blood colligative properties 1 2 3 4  The addition of a solute into water which lowers Tf by 0.52°C will produce an isotonic solution. 1% NaCl lowers Tf by 0.576°C. Therefore, to lower Tf by 0.52°C need 1 x 0.52 % 0.576 = 0.9% NaCl Isotonic solutions have Tf equivalent to blood Tf Fact return to menu  Preparation of an isotonic solution For potent drugs (need very low dose), it is impossible to add the amount which lowers Tf by 0.52°C. Overdose might occur. Therefore, have to use these drugs at low quantity according to the dose specified. But, we may add non-active material to get the required Tf. W = 0.52 - a b W – weight of non-active additive to be added a – weight of the drug x ΔTf of 1% drug in solution b - ΔTf of 1% additive 1 2 3 4  Increase in osmotic pressure Van’t Hoff equation  = inRT V  – osmotic pressure i – Van’t Hoff factor n – moles of solute R – gas constant T – absolute temperature V – volume Osmotic pressure of blood at 273K = 6.72 atm Therefore, injections are isoosomotic or isotonic to blood if their osmotic pressure = 6.72 atm 1 2 3 4  Osmolality and osmolarity The experimentally derived osmotic pressure is frequently expressed as the osmolality, , which is the mass of solute that when dissolved in 1 kg of water will exert an osmotic pressure, , equal to that exerted by a mole of an ideal unionized substance dissolved in 1 kg of water. The unit of osmolality is the osmole (abbreviated as osmol), which is the amount of substance that dissociates in solution to form one mole of osmotically active particles. Thus, 1 mole of glucose (not ionized) forms 1 osmole of solute, whereas 1 mole of NaCl forms 2 osmoles (1 mole of Na+ and 1 mole of Cl-). In practical terms, this means that a 1 molal solution of NaCl will have (approximately) twice the osmolality (osmotic pressure) as a 1 molal solution of glucose. Osmolarity is defined as the mass of solute which, when dissolved in 1 liter of solution, will exert an osmotic pressure equal to that exerted by a mole of an ideal unionized substance dissolved in 1 liter of solution. 1 2 3 4   = MRT M=  RT = 6.72 0.0821 x 273 = 0.3 mol/L = 300 mmolL-1 Solutions will have osmotic pressure = 6.72 atm if concentration of solute(s) or component(s) = 300 mmolL-1 or 300 mosmolL-1 Therefore, if the preparation contains several substances, total concentration of all substances must = 300 mmolL-1 or mosmolL-1 Isotonic solutions have  equivalent to blood  Fact return to menu  Clinical relevance of osmotic effects It is important to ensure that the effective osmotic pressure of a solution for injection is approximately the same as that of blood serum. This effective osmotic pressure, which is termed the tonicity, is not always identical to the osmolality because it is concerned only with those solutes in solution that can exert an effect on the passage of water through the biological membrane. Solutions that have the same tonicity as blood serum are said to be isotonic with blood. Solutions with a higher tonicity are hypertonic and those with a lower tonicity are termed hypotonic solutions. Similarly, in order to avoid discomfort on administration of solutions to the delicate membranes of the body, such as the eyes, these solutions are made isotonic with the relevant tissues. 1 2 3 4 Tonicities (osmolalities) of intravenous fluids The osmotic pressure of many of the commonly used IV products are in excess of that of plasma (291 mosmol dm-3). It is generally recommended that any fluid with an osmotic pressure above 550 mosmol dm-3 should not be infused rapidly as this would increase the incidence of venous damage. The rapid infusion of marginally hypertonic solutions (300-500 mosmol dm-3) would appear to be clinically practicable; the higher the osmotic pressure of the solution within this range, the slower should be its rate of infusion to avoid damage. Patients with centrally inserted lines are not normally affected by limits on tonicity as infusion is normally slow and dilution is rapid. 1 2 3 4 Concept check State true or false The osmotic pressure: a. is a colligative property. True b. allows us to determine the molecular True weight of the macromolecule. c. is proportional to the concentration True of solute. d. could be calculated using the Van’t True Hoff equation.  Decrease in vapour pressure Addition of solute (solid) results in decrease of vapour pressure of liquid (solvent) Vaporization of volatile liquid decreases (effect equivalent to increase of Tb) Therefore, useful to reduce loss of volatile ingredients during processing or mixing especially when using heat to facilitate mixing. 1 2 3 4 Which colligative property is best to calculate MW of solute? Out of the 4 colligative properties, osmotic pressure is the best colligative property to calculate molar mass of solute especially polymers, proteins etc. Because : – Osmotic pressure can be measured at room temperature – Tf and Tb are very small values and difficult to measure with ordinary thermometer 0.5 atm is equivalent to > 15 ft (~5m) return to height in a column of water menu  Colligative properties give good results under the conditions that: The The The The solution is solution is solution is solution is dilute non- non- ideal (obey volatile electrolyte Raoult’s Law) Review questions (fill in the blanks) 1. The sum of the mole fractions of all the components in a solution is 1 _________. 2. 3 g of a substance (MW 30) is dissolved in 250 g of water. The 3/30 x 1/0.25 = 0.4 m/kg molality of the solution is ______________________. 3. The freezing point depression constant depends on the solvent characteristics of ___________ only. 4. K.kg.mol-1 The S.I. unit of boiling point elevation constant Kb is __________. 5. Relative lowering of vapour pressure is equal to the mole fraction of a solute __________ in the solution. 6. The Van’t Hoff factor for phenol which dimerizes in benzene is 0.5 ____________. Review questions (true or false) 1. Solutions having the same osmotic pressure are called isotonic solution. True 2. The value of Kf for water is smaller than its Kb. False 3. For a dilute solution, mole fraction of a solute is directly proportional to True the molality of the solution. 4. The Van’t Hoff factor of Ca(NO3)2 when it is completely dissociated is 1/3. False 5. Colligative properties do not depend upon the amount of solute in the False solution. 6. The correct order of increasing vapour pressure of 0.1 m aqueous solution of glucose, NaCl and magnesium chloride is glucose < NaCl < False magnesium chloride. VAPOUR PRESSURE, RELATED LAWS AND EQUATIONS, TYPES OF SOLUTIONS AND DISTILLATION  RAOULT’S LAW Vapour pressure of a solution is the total of partial vapour pressures of its components which are directly proportional to their mole fractions For a 2 component solution produced by mixing A and B P =P +P S A B Partial vapour Vapour pressure Partial vapour pressure of B of the solution pressure of A PA = P°A. XA P° → vapour pressure of pure substance PB = P°B. XB X → mole fraction of XA + X B = 1 each substance Therefore, PS = P°A. XA + P°B. XB = P°A. XA + P°B(1- XA) = P°A(1- XB) + P°B. XB The equation is applicable when both components are liquids and able to vaporize and have vapour pressure RELATIONSHIP OF VAPOUR PRESSURE WITH MOLE FRACTION FOR SOLUTION CONTAINING TWO LIQUIDS P P PS P°B PB P°A PA 0 1A B1 0 At any mole fraction of A and B, PS = PA + PB  But if the components consist of one liquid and one solid PS = Psolid + Pliquid = P°solid. Xsolid + P°liquid. Xliquid = 0 because most solids do not produce vapour at normal atmosphere Therefore, PS = 0 + P°liquid. Xliquid = P°liquid (1 – Xsolid) known as Henry equation ↑ Xsolid → ↓ Xliquid → ↓ Pliquid → ↓ Psolution RELATIONSHIP BETWEEN PRESSURE OF SOLUTION AND MOLE FRACTION FOR SOLUTION CONTAINING ONE LIQUID AND ONE SOLID P P P°liquid PS = Pliquid 0 1 Xliquid Xsolid 1 0 ↑ Xsolid → ↓ Pliquid → ↓ Psolution Less volatile and more difficult to boil Therefore, Tb ↑ when Xsolid ↑ USES OF RAOULT’S LAW To determine compositions of liquefied propellants in aerosol To classify types of solutions and to predict their properties USING RAOULT’S LAW TO DETERMINE THE COMPOSITIONS OF LIQUEFIED PROPELLANTS IN AN AEROSOL Liquid propellants are very volatile at room temperature because their boiling temperatures are very much lower Lots of vapour is produced and trapped in the aerosol container, this creates pressure which is very much higher than atmospheric pressure and is continuously pressing the aerosol formulation inside When the actuator is depressed, the valve is opened and the liquid preparation is forced out as a spray When actuator is depressed, the valve is opened, the liquid rushes through the opening into the air EFFECT OF PROPELLANT(S) VAPOUR PRESSURE Propellant’s Rate of Properties of Useful as pressure discharge spray Coarse and Surface spray e.g. analgesic, Low Slow wet droplets deodorant sprays Smaller and Medium Fast Insecticide sprays drier droplets Very fine Able to float in the air for spray High Very fast longer duration e.g. air droplets and freshener and deodorizer very dry EXAMPLES OF PROPELLANTS (CFCS) Molecular Trade weight Vapour pressure Chemical name Formula Tb (°C) name (x105Nm-2) Trichloromonofl Freon 11 CCl3F 137.4 23.7 0.93 uoromethane Dichlorodifluoro Freon 12 CCl2F2 120.9 -29.8 5.85 methane Dichlorotetraflu Freon 114 CClF2CClF2 170.9 3.6 1.90 oroethane Monochloropent Freon 115 CClF2CF3 154.5 -38.7 8.10 afluoroethane DETERMINATION OF COMPOSITION OF PROPELLANT MIXTURE TO PRODUCE AEROSOL WITH INTERMEDIATE VAPOUR PRESSURE E.g. to produce propellant mixture having vapour pressure of 3.5 x 105 Nm-2 Can mix – a) Freon 11 (P° = 0.93) with Freon 12 (P° = 5.85) – or b) Freon 11 with Freon 115 (P° = 8.1) – or c) Freon 114 (P° = 1.90) with Freon 12 – or d) Freon 114 with Freon 115 How to determine propellant composition? Graph: mole fraction Calculation using versus vapour pressure Raoult’s law for each component and for solution BY CALCULATION USING RAOULT’S LAW PS = P°A. XA + P°B. XB Example: In a 2-compartment aerosol propellant system (A&B), the blend of two propellants (A/B = 30:70[g]) where pure propellant A (MW 120.93 g/mole) has vapour pressure of 84.9 psia and propellant B (MW 137.38 g/mole) has vapour pressure of 13.4 psia. Calculate the partial pressure for A and B and total vapour pressure of the propellants in the aerosol. BY DRAWING THE GRAPH: MOLE FRACTION VERSUS VAPOUR PRESSURE FOR EACH COMPONENT AND FOR SOLUTION P°Freon12 = 5.85 PS PFreon 12 P P (x 105 Nm-2) (x 105 Nm-2) PFreon 11 P°Freon 11 = 0.93 0 1 Freon 11 X Freon 12 1 0 Based on the required vapour pressure, Ps, we can determine X Freon 11 and X Freon 12 XF11 x MWF11 = wtF11 Convert proportionally to the XF12 x MWF12 = wtF12 weight of propellant mixture Total = _____ to be used  USING RAOULT’S LAW TO CLASSIFY TYPES OF SOLUTIONS AND TO PREDICT THEIR PROPERTIES Liquid and Liquid solutions are divided into – Ideal solutions i.e. solutions which follow Raoult’s Law i.e. solutions having calculated Ps = measured Ps – Non-ideal solutions (true solutions) because they are homogeneous and are in one phase just as ideal solutions, but they do not follow Raoult’s Law i.e. calculated Ps ≠ measured Ps. Measured Ps deviates either Positively or Negatively  By comparing the calculated graph and measured vapour pressures at various X, the solutions can be classified as ideal, negatively deviated or positively deviated solutions max P Calculated PS Measured PS for P°B positively deviated solution Measured PS for P°A Measured PS negatively deviated for ideal solution min P solution 0 1B X A1 0 Azeotrope for Negatively deviated solution has a composition with min PS Positively deviated solution has a composition with max PS Properties or characteristics of Types of solution solution Ideal True Positively deviated Negatively deviated a) Psolution : measured vs Equal Higher Lower calculated b) Total of cohesive forces Equal Adhesive (new) < Adhesive > cohesive between solute-solute solvent- No change in cohesive Difficult to form solvent before dissolution strength Easier to form vapour vs Able to evaporate vapour Pmeasured < Pcalculated Total adhesive forces between as predicted Pmeasured > Pcalculated the newly formed solute- Pmeasured = Pcalculated solvent bond c) Heat absorbed during Heat absorbed ≈ Heat absorbed > Heat absorbed < heat breakage of cohesive bonds heat released heat released released vs Heat released during formation of adhesive bonds d) Heat of solution value ≈0 +ve -ve e) Process of solution - Endothermic Exothermic Properties or characteristics of Types of solution solution Ideal True Positively deviated Negatively deviated f) Temperature of solution Equal to original Decreased Increased formed components g) Volume of solution formed Equal Larger (adhesive < Smaller (adhesive > vs Ʃ of components cohesive. Therefore, cohesive. Therefore, loosely packed) closely packed) h) Presence of azeotrope None At composition with At composition with max PS (easiest to min PS (most difficult vaporise). Therefore, to vaporise). Tb lowest. Therefore, Tb highest. i) Examples i) Chloroform i) Ethyl alcohol + i) HCl + H2O +benzene benzene ii) Acetone + ii) Toluene + ii) Chloroform + ethyl chloroform benzene alcohol iii) Freon _ + freon _  Review questions 1. For a non ideal solution exhibiting negative deviations from Raoult’s law, negative ΔHmix has a ___________ non-zero value. 2. An azeotropic solution of a positively deviated solution boils at a lower ___________ temperature. P increase, T decrease! 3. Non-ideal solutions always show higher vapour pressure than ideal solutions. (T/F) False  Review questions 4. One mole of non-volatile solute is dissolved in 2 mole of water, the vapour pressure of the solution relative to that of water is a) 2/3 b) 1/3 c) 1/2 d) 3/2 5. At 40°C, the vapour pressure of pure liquid benzene and toluene are 160 mmHg and 60 mmHg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids will be a) 140 mmHg b) 120 mmHg c) 110 mmHg d) 100 mmHg Review questions 6. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g is added to 39.0 g of benzene (MW 78 g/mole). The vapour pressure of the solution then is 0.845 bar. What is the molecular mass of the solid substance? P°B = 0.850 bar, MWB = 78 g/mole PS = P°B. XB wtB = P°B. MWB_______ wtB + wtX MWB MWX 39.0 0.845 = 0.850. 78________ 39.0 + 0.5 78 MWX Therefore, MWX = 166.7 g/mole Review questions 7. An aqueous solution of 2% non-volatile solute exerts a pressure of 0.996 atm at the boiling point of the solvent. What is the molecular mass of the solute? P°water = 1 atm, MWwater = 18 g/mole 2% solute = 2g solute + 98g water PS = P°water. Xwater wtwater = P°water. MWwater_______ wtwater + wtsolute MWwater MWsolute 98 0.996 = 1. 18________ 98 + 2__ 18 MWsolute Therefore, MWsolute = 91.7 g/mole DISTILLATION A process in which a liquid or vapour mixture of two or more substances with different boiling points is separated into its component fractions by the application and removal of heat. Distillation Distillation: Solid in Liquid Solution – Simple distillation At normal atmospheric pressure, solids do not vaporize and have no vapour pressure Only liquids vaporize and have vapour pressure. Therefore, liquids can be separated from solids by distillation Tb of solution increases as the concentration of solute increases when amount of liquid decreases during distillation Separation using normal/ simple distillation Distillation: Solid in Liquid Solution – Distillation using rotary evaporator and vacuum at lower temperature Suitable for solution containing thermolabile materials When flask rotate, thin film is formed with large area for evaporation Vacuum pump lowers the pressure inside the system Both help to lower the Tb Distillation using rotary evaporator and vacuum at lower temperature  Distillation: Liquid in Liquid Solution Ideal solution Relationship of P and X Phase diagram at various T and X Pure A: lowest P° and highest Tb P°B TbA Vapour Pure B: highest P° and lowest Tb L+V Simple distillation of e.g. M (XA ≈ 80%, P°A XB ≈ 20%) TbB M Liquid ii) Repeat the process by distilling A B A B Nmixture → Produce 0 X X XA = ↑ i) XB = ↓ i) and ii) The distillates contain more and more A, i.e. the component with higher Tb or lower P° Distillate N M: XA = 0.8 XA < 0.8 XB = 0.2 XB > 0.2 Fractional Distiller As distillates at lower plates vapourize, go up and condense at higher plates, the distillates (fractions) become richer with component of lower Tb and higher P° The fraction at the lower plate becomes richer with component of higher Tb and lower P° It is possible to collect pure substances at topmost and bottommost plates Therefore, for ideal solutions of A and B – Pure B (lowest Tb) can be collected at the topmost plate first – Pure A (highest Tb) will be at the bottommost plate and will come out last Fractional Distiller Distillation: Liquid in Liquid Solution Negatively deviated P vs X solution P°B P°A PAzeotrope < P°A and P°B Pmin Therefore, TbAzeotrope > TbA and TbB Azeotrope Fractional distillation of composition right of A X B Azeotrope: Phase diagram – Pure B come out first V – Azeotrope come out last Fractional distillation of composition left of TbA Azeotrope: L TbB – Pure A come out first Azeotrope – Azeotrope come out last A X B Distillation: Liquid in Liquid Solution Positively deviated Pmax P vs X solution P°A PAzeotrope > P°A and P°B P°B Therefore, TbAzeotrope < TbA and TbB Azeotrope Fractional distillation of composition right of A X B Azeotrope: Phase diagram – Azeotrope come out first TbB – Pure B come out last V Fractional distillation of composition left of TbA Azeotrope: L – Azeotrope come out first Azeotrope – Pure A come out last A X B  Review questions 1. Refer to the figure below. If the mole fraction of B in the liquid is 0.8, what is the composition of B in the vapour above the liquid? What will be the reading on the thermometer at this point? 95°C 0.13 Review questions 2. In distillation, the boiling point is measured: a) Above the surface of the liquid b) In the boiling liquid itself c) In the receiving flask d) With a thermometer external to the distillation apparatus 3. If a compound boils at 100°C at 760 mmHg, its boiling point at 725 mmHg will be: PV = nRT a) Lower P1 = P2 b) Higher T1 T2 c) The same 760 = 725 373 T2 d) Cannot predict Review questions 4. In a distillation of a mixture of two components, the first compound to condense into the receiving flask will be: a) The one with the lower melting point b) The one with the higher melting point c) The one with the lower boiling point d) The one with the higher boiling point 5. Distillation is commonly used in the laboratory as a way to: a) Remove a reaction solvent b) Purify an organic liquid c) Separate the components of a liquid mixture d) All of the above 6. The solution of two liquids which distils without change in composition or temperature is called a) Ideal solution b) Saturated solution c) Isotropic solution d) None of these Ionization, Effect of pH and Methods for Controlling Ionization and pH Why important? Because most of common substances are either Acids Bases or alkalis Salts  Lowry and Bronsted Theory Acid is a substance which donates proton when ionizes HA → A- + H+ acid anion/ salt proton (unionized) (ionized) Base is a substance which accepts proton when ionizes B + H+ → BH+ base proton cation/ salt (unionized) (ionized)  Ionization of ionizable materials or compounds in water a) Strong acids, bases and their salts ~ When dissolve in water will ionize completely (100%) b) Weak acids and bases will ionize when dissolve in water ~Degree or % of ionization depend on their pKa or pKb and pH Unionized form Ionized form c) Salts of weak acids and bases (products of reaction between these compounds with strong base or acid) ~When dissolve in H2O will ionize completely (100%)  Ionization of Weak Acids HA A - + H + 1 (weak acid) (anion) (proton) Conjugate base of Very reactive. weak acid Must change because can into less accept proton reactive form. H+ + H2O H3O+ 2 (water) 1 + 2 HA + H2O A- + H3O+ (acid 1) (base 2) (base 1) (acid 2) * H2O acts as a base  Ionization of Weak Bases H2O OH- + H+ 1 (acid) (proton) Because donate proton Very reactive B + H+ BH + (cation) 2 (weak base) Accepts proton Conjugate acid for weak base 1 + 2 because can donate proton B + H2O BH+ + OH- (base 1) (acid 2) (acid 1) (base 2) *H2O acts as an acid Therefore, H2O is called/ known as an amphiprotic solvent or substance because it can act as acid or base Ionization Constants (pKa and pKb) For weak acid HA A- + H+ Unionized acid Ionized form Ka = [H+][A-] [ ] → concentration [HA] Both sides are log10, log10 Ka = log10 [H+] + log10 [A-] - log10 [HA] Reverse the signs, -log10 Ka = -log10 [H+] - log10 [A-] + log10 [HA] Since -log10 Ka = pKa and -log10 [H+] = pH Therefore, pKa = pH + log10 [HA] - log10 [A-] Thus, pKa = pH + log10 [HA] [A-] Henderson-Hasselbalch equation for weak acid Henderson-Hasselbalch equation for weak acid pKa = pH + log10 [HA] [A-] pH = pKa + log10 [A-] [HA] pH = pKa + log10 [ionized] [unionized] pH = pKa + log10 [salt] [acid] Concept check Acetic acid (CH3COOH) has pKa = 4.8. The pH of a 0.01M solution of this acid is: CH3COOH CH3COO- + H+ a. 3.4 Correct Before 0.01M 0M 0M After (0.01-x)M xM xM b. 4.4 pKa = 4.8 c. 5.4 -log10 Ka = 4.8 Therefore, Ka = 1.58 x 10-5 d. 2.4 By definition, Ka = [H+][CH3COO-] e. 1.9 [CH COOH] 3 So, [H+][CH3COO-] = 1.58 x 10-5 [CH3COOH] x2 = 1.58 x 10-5 0.01-x Assume x is small, x2 = 1.58 x 10-5 0.01 x = 3.98 x 10-4 = [H+] Therefore, pH = -log10 [H+] = -log10 3.98 x 10-4 = 3.4 Henderson-Hasselbalch equation for weak base For weak base B + H+ BH+ Unionized base Ionized form Ka = [H+][B] [BH+] Both sides are log10, log10 Ka = log10 [H+] + log10 [B] - log10 [BH+] Reverse the signs, -log10 Ka = -log10 [H+] - log10 [B] + log10 [BH+] pKa = pH + log10 [BH+] [B] Henderson-Hasselbalch equation for weak base Henderson-Hasselbalch equation for weak base pKa = pH + log10 [BH+] [B] pH = pKa + log10 [B] [BH+] pH = pKa + log10 [unionized] [ionized] pH = pKa + log10 [base] [salt]  Other useful equations i. pKa = pKw – pKb pKw = 14 ii. pKb = -log Kb iii.pKw = -log Kw  Examples of Drugs and Their pKa Acetic acid 4.76 Ascorbic acid 4.2 Acetylsalicylic acid 3.59 Citric acid 3.1 (has 3 COOH) 4.8 3pKa polybasic acid 5.4 Benzocaine 2.78 Ephedrine 9.36 Chlorpheniramine 7.2 Chlorpromazine 9.21 Adrenaline 8.5 for amine group 9.9 for phenol group 2pKa Morphine 8.3 2pKa polyacidic bases 9.5 Tetracycline 3.3 7.2 3pKa 9.5 Compounds with 2 or more pKa are called polyprotic compounds (the bases as polyacidic and the acids polybasic) Relationship between degree of weak acid and weak base ionization and solubility with their pKa and pH of solution If we know the pKa of a weak acid or a weak base, we can calculate the % which is ionized at various pH values by using the Henderson-Hasselbalch equation If the solubilities of the weak acid or the weak base at various pH are measured, the relationship or graph as below can be obtained 100 100 Relative % 50 50 solubility Ionized (%) pH=pKa-2 pH=pKa pH=pKa+2 0 0 pH  For weak acid i. At pH ≤ pKa – 2, solubility lowest 100% unionized, 0% ionized ii. At pH > pKa-2, solubility ↑ % unionized ↓, % ionized ↑ iii. At pH = pKa, relative solubility 50% % unionized = 50%, % ionized = 50% iv. At pH ≥ pKa + 2, solubility maximum % unionized 0%, % ionized 100% For weak base → reverse effects i. At pH ≤ pKa – 2, solubility maximum 100% ionized, 0% unionized iv. At pH ≥ pKa + 2, solubility minimum % unionized 100%, % ionized 0%  Example of salicylic acid (pKa=3) Stomach pH=1.2 Blood pH =7.4 C6H4(OH)COOH C6H4(OH)COOH C6H4(OH)COO- + H+ C6H4(OH)COO- + H+ salt [ A ] salt [ A ] Ratio    10(1.23) Ratio    10( 7.43) acid [ HA] acid [ HA] salt salt  1.58 x10  2  2.51x104 acid acid Drug Gastric side Blood side C6H4(OH)COOH 1 1 C6H4(OH)COO-.0158 25100 Total drug conc. 1.0158 25101  Significance of Relationships Δ ionization Δ Ko/w Δ pH Δ solubility Δ apparent size Δ physical stability (i) Gas from e.g. precipitation atmosphere (ii) Materials from container Δ absorption Δ bioavailability Affects degradation or breakdown of drugs Δ potency i) Hydrolysis ↑ at lower pH ii) Microbial degradation ↑ at higher pH Δ efficacy or effectiveness of drugs Therefore, must prevent change of pH during storage  Buffer Mixture Mixture of substances which reduce (minimize) or prevent change of pH when acid or base is added into the solution Can be prepared by mixing: a) Weak acid and its salt (conjugate base for weak acid) E.g. citric acid + sodium citrate acetic acid + potassium acetate b) Weak base and its salt (conjugate acid for the weak base) E.g. ephedrine and ephedrine HCl * Ratio of weak acid and its salt and ratio of weak base and its salt can be calculated by Henderson-Hasselbalch equation  Mechanism how buffer mixture minimizes or retards change in pH Weak acid and its conjugate base (salt) – If acid is added, H+ of the added acid will ↓ pH significantly but salt or conjugate base will react with H+ to produce weak acid H+ + A- → HA Salt or Weak acid, same as component of buffer. conjugate base Any Δ in pH, very little. – If alkali is added HA + OH- → A- + H2O Weak acid Salt or conjugate base  Weak base and its conjugate acid (salt) – If acid is added B + H+ → BH+ Weak base Salt or conjugate acid, the same or buffer as buffer component. Therefore, Δ in pH minimum. – If alkali is added BH+ + OH- → B + H2O Salt or conjugate Weak base, same as acid of buffer buffer component. Δ in pH retarded or minimized.  Example 1 Calculate the amount of Na.acetate to be added to 100 ml of 0.1 mole.L-1 acetic acid solution to produce a pH 5.2 buffer (pKa acetic acid = 4.76; MW Na.acetate = 82 g/mole). pH = pKa + log10 [salt] [acid] 5.2 = 4.76 + log10 [salt] 0.1 log10 [salt] = 0.44 0.1 [salt] = 2.754 0.1 [salt] = 0.2754 mole.L-1 Therefore, weight of salt to be added = 0.2754 x MW x 0.100 = 0.2754 x 82 x 0.100 = 2.258g  Example 2 Calculate the change in pH when 10 ml of 0.1 mole.L-1 NaOH is added to the buffer solution in Example 1. 10 ml of 0.1 mole.L-1 NaOH = 0.001 mole Therefore, 0.001 mole NaOH reacted with 0.001 mole acetic acid to produce 0.001 mole Na.acetate Therefore, pH of resulting mixture = pKa + log10 [salt] [acid] = 4.76 + log10 0.02754 + 0.001 0.01 – 0.001 = 5.26 Therefore, pH = 5.26 - 5.2 = 0.06 (very minimum, not significant)  Buffer Capacity (strength), β Buffer capacity is the ability of the buffer mixture to resist change in pH when acid or alkali is added The capacity or strength is reflected in the amount of strong alkali or acid (in mole) required to change pH of 1 L buffer by 1 unit i. β = dC If amount of acid or alkali needed to change d(pH) pH by 1 unit is high, it reflects that the buffer is strong or the capacity is high. Concentration of Small amount of acid and alkali won’t change buffer mixture pH much. ii. β = 2.303 CO.Ka.[H3O+] ↑ CO → ↑ β + ([H3O ] + Ka)2 iii. Maximum buffer capacity, βmax = 0.576CO iii. Buffer capacity of specified mixture is highest when pH of buffer solution = pKa of weak acid or base used If pH of buffer solution further from pKa → β ↓ If pH of buffer solution closer to pKa → β ↑  Advantages of adding buffer mixture into aqueous products a) Maintain stability of products i. Δ in pH caused by leaching of alkali from wall of container is prevented or minimized ii. Δ in pH caused by diffusion of atmospheric gas e.g. CO2 through wall of plastic container and space between cap and container is retarded When pH remains constant, changes in ionization, solubility, etc. is retarded. The product behaves as expected b) Ensure effectiveness of drug Some drugs are degraded in the stomach (low pH) e.g. erythromycin. Therefore, buffer is added into the tablet. When it disintegrates and dissolves in the stomach, pH in the stomach remains high (maintained by buffer), drug is stable and is absorbed as intended. c) Reduce irritation and pain Most of product to be applied at sensitive area e.g. eye must have pH ≈ pH of body fluid at site of application. Therefore, can be buffered at those pH. But, if product must be buffered at other pH because of stability, β of buffer must be minimum to ensure physiological buffer can take over. Review questions 1. A solution of pH 9.0 is one thousand times as basic as a solution of pH _______ a) 6 b) 7 c) 4 d) 10 2. An acidic buffer can be prepared by mixing _______ a) Ammonium acetate and acetic acid b) Ammonium chloride and HCl c) H2SO4 and Na2SO4 d) CH3COOH and H2SO4 Review questions 3. The correct operational relationship between pKa and pH is that ________ a) Both are log functions b) Both are always 7 for bases c) When pH = pKa, the ionizable compound in question will have a charge of +0.5 d) When pH = pKa, the ionizable compound in question will be half protonated and half deprotonated 4. Physiological pH is 7.4. What is the hydrogen ion concentration of a solution at physiological pH? a) -7.4 M pH = 7.4 b) 0.6 M -log10 [H+] = 7.4 c) 1 x 10-8 M [H+] = 3.98 x 10-8 d) 4 x 10-8 M ≈ 4 x 10-8 M Review questions 5. What is the molar ratio of salt/base required to adjust the pH to 8.5 using ammonium hydroxide-ammonium chloride buffer, using Kb of ammonium hydroxide of 1.77 x 10-5 at 25°C? pH = pKa + log10 [base] [salt] 8.5 = 14 -(-log10 (1.77 x 10-5)) + log10 [base] [salt] 8.5 = 14 - 4.75 + log10 [base] [salt] log10 [base] = -0.75 [salt] [base] = 0.18 [salt] [base] : [salt] = 0.18 : 1 [salt] : [base] = 1 : 0.18 = 5.56 : 1 Review questions 6. What percentage of atropine base exists as unionized (base form) in ophthalmic drops buffered at pH 6.8, using a pKb of atropine of 4.35 at 25°C? pH = pKa + log10 [base] [salt] 6.8 = 14 – 4.35 + log10 [base] [salt] log10 [base] = -2.85 [salt] [base] = 1.4 x 10-3 = 0.0014 [salt] [base] : [salt] = 0.0014 : 1 % [base] = 0.0014 x 100% 1.0014 = 0.14% Review questions 7. What is the pH of a solution containing 0.09 M pseudotropine solution and 0.3 M pseudotropine sulfate solution in 1 L, using Ka of pseudotropine 6.2 x 10-11 at 25°C? pH = pKa + log10 [base] [salt] = - log10 (6.2 x 10-11) + log10 0.09 0.3 = 10.2 – 0.5 = 9.7 Complexation The formation of a species by association of two or more interacting entities (atoms, ions, molecules or polymers) which is more complex in nature, structure or configuration  Complexes In pharmacy, depending on the interaction involved, we classify them as – Coordination complexes, or – Molecular complexes The interaction may lead to the formation of: – Ionic bond – Covalent bond – Hydrogen bond Single or in – Debye bond (dipole-induced dipole) combination – Keesom bond (dipole-dipole) – London bond (induced dipole-induced dipole or dispersion force) The complex formed is reversible if the bond is weak and conditions favour the reversal. Irreversible if the bonds are strong. Coordination complexes Ion containing central metal cation bonded to one or more molecules or ions, forming coordinate bonds Complex ions are crucial to many chemical and biological processes Metal ion (transition metal) and ligand interaction resembles a Lewis acid-base reaction Ligand is Lewis base which donates electron pair to metal cation to form coordinate covalent bond Example: Co2+ (aq) + 4 Cl- CoCl42- (aq) pink blue Coordination number – the number of nearest neighbours of an atom or an ion in a lattice, or number of species bound to a coordination compound  Simple coordination complexes i. HgI2 + 2KI → (HgI4)2-2K+ Dipole Ionic Very soluble (low solubility) (very soluble) complex Cannot prepare concentrates Bulky volume Large container I I- K+ Problem in transporting Counterion Low acceptability Large storage area Hg I- Ligand K+ Central atom or I (unidentate ligand because each ligand nucleus forms 1 bond with central atom) Uses: ↑ solubility ↑ concentration ↓ bulkiness or volume of product, therefore, more convenient. Can be diluted to the strength needed  Simple coordination complexes NH3 ii. Co(NH3)3 + 3NH3Cl NH3 NH3 low solubility very soluble Co 3Cl- Central atom or nucleus NH3 NH3 NH3 Counterion Hexaammonium cobalt chloride Ligand unidentate (Very soluble complex)  The coordination number - the number of donor atoms bonded to the central metal atom.  Oxidation number or oxidation state – the number of electrons associated with an element in a compound. For example, for S, it is +6 in SO42- and +4 in SO32-  Coordinate bonds between ligands and metals do not affect oxidation state, so the oxidation state for Co in the example is still +3  Coordination complex with chelating or sequestering agents i.e. polydentate agents i. With bidentate ligand E.g. ethylene diamine CHNH2 Each molecule has 2 places for bonding with metal or central atom. CHNH2 Therefore, called bidentate ligand. ii. With tridentate ligand E.g. citric acid HOCHCOOH 3 places where central atom can CHCOOH form bonding with. Therefore, called tridentate ligand. HOCHCOOH  Coordination complex with chelating or sequestering agents i.e. polydentate agents iii. With hexadentate ligand E.g. Ethylenediaminetetraacetic acid (EDTA) Hexadentate because 6 places to form bonds with central atom  Uses of polydentate ligands (chelating and sequestering agents) Chelate with metal ions catalyzing degradation of drugs or natural products e.g. Vitamin C in fruit juices. The chelated metal unable to act as catalyst. Therefore, breakdown of drugs or products is decreased. Stability of product ↑. Chelating agents will form complexes with metal ions inside cells’ organelles especially those in rapidly multiplying cells such as cancer cells. The chelated metal cannot function thus biological process of cells disturbed and cells will be killed. Unfortunately, rapidly multiplying normal cells are effected also. Nickel coordination complexes Molecular Complexes Formed by (weaker) non-covalent interactions between the substrate and ligand: – Hydrogen bond – London bond – Keesom bond Van der Waals interactions – Debye bond – Hydrophobic interactions – interfacial phenomenon that results from attraction between non-polar (hydrophobic) groups with water molecules – Electrostatic interactions – occur between charged amino acids with oppositely charged ligand molecules. Thus, the complexation is reversible Usual functional groups involved: – Ester – Amine – Phenol – Ketone – Ether Examples are – small molecule-small molecule complexes – small molecule-macromolecule complexes (drug-protein and enzyme-substrate)  Small molecule-small molecule complexes Drug-drug complexation definitely alters the pharmaceutical properties of a dosage form, to a small or to a large extent Property Example Chemical stability Benzocaine-caffeine Solubility Caffeine-gentisic acid Dissolution rate Griseofulvin-hexanoic acid Partition coefficient Benzoic acid-caffeine Permeability Prednisolone-dialkylamides Absorption rate Salicylamide-caffeine Pharmacological activity Butesin-picric acid Formulation pharmacists must be aware of possible complex formation between a drug and an excipient to avoid dosage form incompatibilities that may in turn affect patients’ therapy  Complexation with Caffeine Caffeine has a weak dipole and can form complexes with numerous drugs E.g. – Ergotamine (antimigraine) – Benzoates (analgesic) – Salicylates (analgesic) – Barbiturates (hypnotic) – Sulphonamides (antibiotic)  Caffeine with Benzocaine OCH2CH3 - CH3 C O O CH3 N + :N CH -O N N N: + CH3 H2 – Solubility of benzocaine ↑ by caffeine – Polarity of functional groups ↓ thus Ko/w ↑, absorption ↑, effectiveness ↑ and bioavailability ↑ – Functional groups protected from degradation  Caffeine with Ergotamine (Cafergot®) Caffeine increases solubility of er