PF1010 HAM 2425 Lecture and Tutorial Questions (2025) PDF

PF1010 Physicochemical Basis of Pharmaceuticals Dr. Humphrey Moynihan ([email protected]) & Prof. Abina Crean ([email protected])  Material on Canvas Canvas → 2025-PF1010 Module (i.e., Canvas module): Dr. Humphrey Moynihan Materials Will contain:...

PF1010 Physicochemical Basis of Pharmaceuticals Dr. Humphrey Moynihan ([email protected]) & Prof. Abina Crean ([email protected])  Material on Canvas Canvas → 2025-PF1010 Module (i.e., Canvas module): Dr. Humphrey Moynihan Materials Will contain: – PDFs of lecture slides and tutorial question slides – PDFs of tutorial answers – Recordings of lectures if possible  PF1010 Objective Module objective: Introduction to the physicochemical principles of pharmaceutical systems. Pharmaceutical systems, examples include: Formulations of APIs and excipients – API: Active Pharmaceutical Ingredient, i.e., active drug molecule – Excipients: other non-active ingredients of the medicine Drugs binding to receptors in tissues – Receptors: proteins which act as binding sites for drugs Enzymes binding substrates Cellular structures, etc.  PF1010 Objective Module objective: Introduction to the physicochemical principles of pharmaceutical systems. Physicochemical principles: any aspect of: Chemistry Physics or physical chemistry which… …determines or affects the structure, stability and function of pharmaceutical systems.  Textbooks Entry level – Any 3rd level general chemistry textbook (e.g: ‘Chemistry’, C. E. Housecroft & E. C. Constable) – Principles and problems in physical chemistry for biochemists, Price & Dwek Intermediate – Pharmaceutics; The Science of Dosage Form Design, Aulton (Ed.) Advanced – Physical Pharmacy and Pharmaceutical Science, Martin – Physical Chemistry, Atkins & de Paula  H. Moynihan PF1010 Topics Introduction to thermodynamics: equilibrium, ideal gases HM First law of thermodynamics: enthalpy, thermochemistry HM Second law of thermodynamics: entropy, free energy HM Free energy and equilibrium HM Chemical potential, the phase rule HM Phase diagrams, triple and critical points HM Systems of one, two, three components, eutectic points, , triangular phase HM diagrams Systems with multiple solid phases; polymorphs and solvates HM Acids & bases; pH, pKa; Henderson-Hasselbalch equations HM Activity and ionic strength HM i.e. thermodynamics (study of the energy of physical and chemical systems) Thermodynamics and pharmaceuticals A dosage formulation (medicine) is a system of multiple phases and components governed by thermodynamics and process equilibria System: part of the physical world defined for study Phase: homogenous portion of physical material bounded by interfaces Component: chemical ‘ingredient’ of the system [Pharmaceuticals: chemical components of medicines] Thermodynamics and pharmaceuticals A dosage formulation (medicine) is a system of multiple phases and components governed by thermodynamics and process equilibria Examples of equilibrium processes: – Binding of drugs to receptors or enzymes – Biochemical reactions in body metabolism – Processes for manufacturing Active Pharmaceutical Ingredients (APIs) – Many formulation processes – Most measures of pharmacological activity are, effectively, equilibrium constants Thermodynamics and pharmaceuticals A dosage formulation (medicine) is a system of multiple phases and components governed by thermodynamics and process equilibria Pharmaceutical analysis relies on quantitative measurement of partitioning between phases – chromatography, e.g., gas chromatography, TLC, HPLC or enthalpy (heat) transfers – Thermal Analysis or other quantitative physical effects  System and Surroundings System: defined part of the physical world under study Surroundings: rest of the physical world (or at least that part affected by changes to the system) Examples: a dosage form (tablet, suspension), a reaction in a vessel, a biochemical system  Equilibrium Pharmaceutical systems may consists of many components and phases, e.g. Components: API, excipients Phases: solid phase, immiscible liquids, gaseous phases Components may be partitioned between various phases System (phases & components) exist in dynamic equilibrium  Law of mass action & equilibrium constants E.g., say substances A and B react to form substances C and D Say the process requires a ratio of A to B of a:b and gives a ratio of C to D of c:d (a, b, c and d are known as the reaction stoichiometries) Can summarise the process as follows: The symbol means that the process occurs in both directions  Law of mass action & equilibrium constants React a of A with b of B C and D start forming (in a ratio c:d) C and D can also react to form A and B Eventually, the process ‘settles down’ to give constant amounts of A, B, C and D The process is then at equilibrium Is a dynamic equilibrium i.e., although the overall amounts of A, B, C and D stay the same, the process is still on-going in both directions  Law of mass action & equilibrium constants Equilibrium processes subject to the law of mass action E.g., for Can define the equilibrium constant (Keq) for the process as follows E.g., [C] stands for the concentration of component C Note the role played by the stoichiometries a, b, c and d in the equilibrium constant (Will offer a justification for this equation later) Examples of pharmaceutical processes and equilibria Drug partitioned between an aqueous medium (e.g. cell interior) and a lipid medium (e.g. cell membrane) Define drug concentration in both: [drug]aq, [drug]lipid The following equilibrium and equilibrium constant exist: 𝑑𝑟𝑢𝑔 𝑙𝑖𝑝𝑖𝑑 𝐾𝑒𝑞 =𝑃= 𝑑𝑟𝑢𝑔 𝑎𝑞. [known as a partition coefficient (P); generally use log P]  Reaction equilibria 𝑅𝑆𝑆𝑅 [𝐻2ሿ 𝐾𝑒𝑞 = 𝑅𝑆𝐻 2 Note: for reverse process 1 Equilibrium constant is inverted 𝐾𝑒𝑞 𝑙𝑒𝑓𝑡 → 𝑟𝑖𝑔ℎ𝑡 = 𝐾𝑒𝑞 𝑟𝑖𝑔ℎ𝑡 → 𝑙𝑒𝑓𝑡  Biochemical equilibrium: worked problem The concentration of a solution of G3P was initially 0.05 M. Isomerase was added. After the mixture came to equilibrium at 25 °C, the concentration of G3P was 0.002 M. Calculate Keq for the reaction at 25 °C. ‘M’ is short for moles per litre (mol L−1) Moles quantify the amount of a substance, will explain later. Define the equilibrium and the equilibrium constant [DHAP] Keq = [G3P] [G3P] = 0.002 M [DHAP] = 0.05 M − 0.002 M = 0.048 M 0.048 M Keq (at 25 ◦C) = = 24 0.002 M  Tutorial. Equilibrium Constants Keq (at 25 ◦C) for isomerisation of glucose-6- phosphate (G6P) to fructose-6-phosphate (F6P) is 0.428. Define the equilibrium constant Calculate [F6P] at equilibrium at 25 ◦C if the initial [G6P] is 0.1 M  Structure and stability of pharmaceutical systems Pharmaceutical systems: – E.g: medicines (dosage formulations of APIs and excipients) – Ligands & receptors, enzymes & substrates/inhibitors, others Chemical components: APIs, excipients, biomolecules Phases: solids, liquids, gases, other? Key issue: structure and stability  Structure and stability of pharmaceutical systems Structure – Equilibrium constants – Dynamic equilibria Stability – System at or not at equilibrium? – System and surrounding respond to reach equilibrium The energy of the system is the key determinant of equilibrium, structure and stability  Energy and equilibrium Need to find the relationship between the energy of a system and the equilibrium constant K for any process First need to clarify what is meant by the energy of a system Pharmaceutical systems very complex; to develop theory better to start from the simplest systems that can be imagined Theory can subsequently be modified to allow for more complex systems  Ideal gases Simplest possible systems: molecules moving at random through space not interacting No chemical interaction between them No defined spatial relationships Hence no need to allow for the energy contributions of interactions between molecules Such systems are known as ideal gases [also assumed that the molecules in ideal gases occupy no space] Some gases behave close to ideal under certain conditions  Studies on ideal gases Studies on the variation of pressure (P), volume (V) and temperature (T) of a fixed amount of an ideal gas Boyles’s law: P vs. V at constant T Charles’ law: V vs. T at constant P Allowed proposal of the ideal gas law  Boyle’s law and Charles’ law ‘∝’ means ‘proportional to’ Robert Boyle (1627-1691) Jacques Charles (1746-1823) 1 𝑃∝ 𝑉∝𝑇 𝑉 at constant P at constant T  Pressure (P), volume (V) and temperature (T) studies on fixed amounts of (assumed ideal) gases 1 Boyle’s law 𝑃∝ at constant T 𝑉 Charles’ law 𝑉∝𝑇 at constant P 𝑃𝑉 Combining gives: = constant 𝑇 For a fixed quantity of ideal gas  Avogadro’s Hypothesis (1811) Samples of different gases at the same pressure (P), volume (V) and temperature (T) contain the same amount of substance Could be generalised to allow a definition of unit of amount of substance: the mole (mol) Definition: 12.0 g of pure 12C isotope contains one mole of carbon atoms Contains 6.022 x 1023 mol−1 of atoms, known as Avogadro’s number Note: is a vast number 1 mol of any pure chemical substance contains Avogadro’s number of entities The quantity (n) of substance is usually given as the number of moles  Relationship between pressure (P), volume (V) and temperature (T) of n moles of an ideal gas 𝑃𝑉 The gas constant = constant = 𝑅 = 8.314 J mol−1 K−1 𝑛𝑇 (value of R depends on the units) PV = nRT the ideal gas law Note: T in Kelvin (K) [0 °C = 273.15 K]  Mixtures of ideal gases Mixture of two ideal gases, A and B, total pressure (P) nA moles of gas A; nB moles of gas B 𝑛𝐴 + 𝑛𝐵 𝑅𝑇 𝑃= 𝑉 PA and PB are known as the partial pressures of A and B  Mole fraction For component A, for example: xA is the mole fraction of component A In general, mole fraction of component i of any mixture is ‘∑’ means the sum of all the terms ni (i.e., the sums of the numbers of moles of all components of the mixture)  PV = nRT problem A 1L cylinder contains 5.00 g of a gaseous anaesthetic (MW 42.1 g mol−1). The cylinder will leak if the pressure exceeds 1.0 x 106 Pa. From what temperature (in ◦C) will the cylinder leak? R = 8.314 Pa m3 K−1 mol−1 1L = 10−3 m3  Tutorial If the mole fraction of oxygen (O2) in air is 0.22, what is the mass of oxygen in 1.0 L of air at 20 °C and 1.0 atm (atmosphere) pressure, assuming ideal behaviour? Use R = 0.082 L atm mol−1 K−1 Need PV = nRT, but solving for n Need to convert °C to K Need Molecular Mass of O2 (32 g mol−1 )  First Law of Thermodynamics “Energy can neither be created nor destroyed” Based on observation and experience Internal energy (U) – An assembly of atoms, ions and/or molecules constitute the system – For an ideal gas, energy due to: translational, rotational and vibrational motion; ‘storage’ of electrons – (No energy due to interactions between molecules of the gas) “The internal energy, U, of an isolated system is constant”  Changes to the internal energy of a system If system in contact with surroundings, when changes in U occur U = Uafter change − Ubefore change] Usystem + Usurroundings = 0 (first law) i.e. Usystem = −Usurroundings Changes to the internal energy of a system Several ways of changing the internal energy of a system But assuming no changes to the phases or components of the system… – (i.e., no reactions generating new components, no processes, such as crystallisation, generating new phases) …energy of the system can only change through: Heat being transferred to/from the system, or Work being done on/by the system Changes to the internal energy of a system Heat and work both forms of energy The amount of energy transferred through heat exchange quantified by q The amount of energy transferred through work being on quantifed by w Leads to another formulation of the First Law: U = q + w  U = q + w Work done on the system (+ve) Change in internal Heat absorbed by the system (+ve) energy of the system ▪Heat: energy transferred between system and surroundings as a consequence of temperature differences between system and surroundings ▪Quantified by value of q ▪Work: energy transferred between system and surroundings capable, in principle, of lifting a weight ▪Quantified by value of w  Relating work (w) to P and V Imagine an ideal gas in a cylinder trapped by a piston Imagine an infinitely small volume change dV Pressure P unchanged Volume change dV at constant pressure P Expansion; work done by the system (-ve) Work done = force × distance = pressure × change in volume, i.e. PdV Expansion of gas from volume Vi to volume Vf at constant pressure P Expansion; work (w) done by the system (-ve) System reaches equilibrium after each infinitismal change dV Work done = force × distance = pressure × change in volume The total work done the summation of all the PdV terms  U = q + w U = q − PV At constant volume, V = 0, so U = qv (subscript ‘v’ implies constant volume ) Most pharmaceutical processes occur at constant pressure U = qp − PV (subscript ‘p’ implies constant pressure)  Enthalpy H U = qp − PV i.e. qp = U + PV Define a new energy function Enthalpy (H) H = U + PV  Enthalpy H Enthalpy H = U + PV For a process at constant pressure (Hf − Hi) = (Uf −Ui) + P(Vf − Vi) H = U + PV [H = qp] The change in enthalpy equals the heat transferred at constant pressure  Thermochemistry Examines heat transfers (enthalpy changes) during several important process E.g. melting of drug substance, binding of drug to receptor, dilution of drug solution, reaction of precursors to form drug substance, etc. E.g. enthalpy of fusion (melting), enthalpy of binding, etc. Enthalpy of formation (Hf): heat absorbed at constant pressure when 1 mole of a compound is formed (at some specified temperature) from its element in their most stable forms Hf⁰ : value of Hf at 25 ⁰C and 1 atmosphere pressure Heat lost from system to surroundings: exothermic process (H -ve) Heat gained by system from surroundings: endothermic process (H +ve)  Thermodynamic state functions The value of a state function depends only on the present condition (state) of the system and not on the history of the system Independent of the pathway by which the present state was obtained U and H are thermodynamic state functions w and q are not (are pathway dependent)  Intensive vs. extensive properties Intensive properties: independent of the size of the system. E.g. pressure (P), temperature (T) Extensive properties: dependent on the size of the system E.g. internal energy (U), enthalpy (H). [units: kJ mol−1] Is the basis of thermal analysis  Hess’s Law “If a process is carried out in several steps, H for the process will equal the sum of the enthalpy changes for the individual steps” E.g. determining Hf⁰ for CS2 (l) given Hf⁰ CS2 (l) = H1⁰ + 2 H2⁰ − H3⁰ Hf⁰ CS2 (l) = (−393.3) + 2(−292.9) − (−1108.8) kJ mol−1 = 129.7 kJ mol−1  Enthalpies of reaction Hrxn = ∑Hf° (products) − ∑ Hf° (reactants) At a specified temperature (298 K in the following example) E.g. C2H4 (g) + Cl2 (g) → C2H4Cl2 (l) Given: Hf° (C2H4 (g) ) 52.5 kJ mol−1, Hf° (C2H4Cl2(l) ) −167.0 kJ mol−1, Hf° (Cl2 (g) ) 0 kJ mol−1 (element) [Hf° for an element in its standard state = 0] Calculate Hrxn Hrxn = (−167.0 ) − [52.5 + 0] = −219.5 kJ mol−1 Tutorial. Thermochemistry problems A. Given the enthalpies of formation (at 298 K) of fumaric acid and maleic acid are −807.6 and −784.4 kJ mol−1 respectively, determine Hrxn for the following process. maleic acid → fumaric acid (at 298 K) B. Glucose (C6H12O6) is metabolised in the presence of oxygen (O2) to give carbon dioxide (CO2) and water (H2O). (i) Write a balanced equation for the reaction of glucose and oxygen to give carbon dioxide and water (ii) Using the following data (305 K), calculate the enthalpy of the process. Hf⁰ C6H12O6 (s) = −1273.0 kJ mol−1 Hf⁰ O2 (g) = 0 kJ mol−1 (element) Hf⁰ H2O (l) = −285.8 kJ mol−1 Hf⁰ CO2 (g) = −393.5 kJ mol−1  The direction of spontaneous change The first law of thermodynamics describes the energy balance in a given process (Key parameter is the change in enthalpy ΔH) But it does not indicate the preferred direction of the process i.e., the direction of spontaneous change To predict the position of equilibrium for a process, need ΔH, but also need to know the direction of spontaneous change  Spontaneous processes By observation: many processes only occur in one direction – Heat flows from a hotter body to a colder body, not vice versa – Gas molecules initially separated mix randomly but not the reverse, i.e.: Partition removed Reverse process not observed spontaneously Process can occur in the reverse direction but requires intervention  Entropy Entropy (S): degree of disorder or randomness of a system Higher S implies greater disorder; lower S greater order (or organisation) Entropy is a thermodynamic state function The change in entropy (ΔS) for a process is the key parameter for determining the direction of spontaneous change Entropy (S) vs. temperature for a single component Note entropy decreases as temperature decreases i.e. going from less ordered to more ordered phases Large drops in entropy at phase transitions Within phases, some decrease in entropy (increasing order) with decreasing temperature Entropy is a thermodynamic state function S for a particular process is independent of path S at 0 K = 0 J mol−1 K−1 (3rd Law of Thermodynamics) Energy; relating temperature, heat and entropy Any form of energy can by described as an intensity factor × a capacity factor The intensity factor is the driving force The capacity factor is the extent over which the force is conveyed E.g., for mechanical energy: force (intensity) × distance (capacity) or pressure (intensity) × volume (capacity) For electrical energy: potential (intensity) × quantity of charge (capacity)  Heat (q) as a form of energy Likewise can be described as an intensity factor multiplied by a capacity factor The intensity factor is the temperature (T) The capacity factor is the entropy (S) Hence: q = T × S For a given process, the change in entropy is given by S 𝑞 ∆𝑆 = 𝑇  Reversible and irreversible processes Spontaneous processes: system and surroundings not at equilibrium Process continues until equilibrium reached: irreversible process Reversible process: system and surroundings at equilibrium Dynamic equilibrium [system and surroundings at equilibrium after each infinitesimal change]  Reversible transfer of heat E.g. ice and water in equilibrium at 0 ◦C Temperature of ice and water in contact both at 273.15 K during melting or freezing Hmelting = 6.01 kJ mol−1 (equals the amount of heat q absorbed from the surroundings when 1 mol of ice melts reversibly) (H = q for a process at constant pressure) 𝑞𝑟𝑒𝑣 ∆𝑆 = = (6.01 kJ mol−1)/(273.15 K) = 22.0 J mol−1 K−1 𝑇 Heat gained by the ice equals the heat lost by the surrounding water, so S for both the system and surroundings are equal No overall change in entropy of the universe  Irreversible transfer of heat Transfer of heat from a hot object (Thot) to a cold object (Tcold) Hot object loses heat (−q), cold object gains heat (q) 𝑞 −𝑞 ∆𝑆𝑐𝑜𝑙𝑑 = ∆𝑆ℎ𝑜𝑡 = 𝑇𝑐𝑜𝑙𝑑 𝑇ℎ𝑜𝑡 Thot > Tcold, so 𝑞 −𝑞 ∆𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = ∆𝑆𝑐𝑜𝑙𝑑 + ∆𝑆ℎ𝑜𝑡 = + = +𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑇𝑐𝑜𝑙𝑑 𝑇ℎ𝑜𝑡  Second law of thermodynamics 𝑞𝑟𝑒𝑣 Reversible process: ∆𝑆 = 𝑇 𝑞𝑖𝑟𝑟𝑒𝑣 Irreversible process: ∆𝑆 > 𝑇 𝑞 ∆𝑆 ≥ 𝑇 This is the second law of thermodynamics In words: “In a spontaneous process, the entropy of the universe increases” Energy and equilibrium of systems, so far… From the 1st law: enthalpy H = qp= U + PV Concerns energy changes to the system during processes; lowering enthalpy preferred From the 2nd law: entropy S = q Concerns changes to the disorder of the system during processes; increasing entropy preferred Need a thermodynamic concept which captures both of these  Criterion for the direction of spontaneous change For any spontaneous process at constant T and P A spontaneous process implies irreversible change, i.e: 𝑞𝑖𝑟𝑟𝑒𝑣 ∆𝑆 > or qirrev < TS 𝑇 q = H (const. T, P) So H < TS or H − TS < 0 Process will continue as long as H − TS < 0 (Gibbs) Free Energy Require: H − TS < 0 Define G = H − TS J. Willard Gibbs (1839-1903) Process will proceed as long G < 0 G known as the Gibbs free energy G is a thermodynamic state function  G = H − TS If G < 0, preferred direction of process Arrives at equilibrium when G = 0 If the H term dominates, have an enthalpy controlled process (H large and –ve) If the TS terms dominates, have an entropy controlled process  Enthalpy vs. entropy E.g. analysis of Mg2+ using complexation by EDTA EDTA tetra sodium salt Widely used method for analysis of Mg ions in aqueous solution Note: Mg2+ ion highly solvated by water Method used pharmaceutically, in pharmacology, biochemistry, etc. Note, process goes essentially to completion i.e., fully to the right hand side as drawn above  Enthalpy vs. entropy Can determine the relative contributions of enthalpy and entropy to the process. Hence can determine the thermodynamic driving forces for the process and… …infer the molecule-level processes Find that H small and +ve But that S is significant and +ve Release of five H2O molecules and two NaCl, increased disorder (more components generated) Overall G –ve, process spontaneous left to right as drawn  Standard states of free energies G°: value of G for a pure substance under 1 atmosphere of pressure (at some specified temperature) G°: change in free energy when 1 mole of products in their standard states are converted to 1 mole of products in their standard states  Using ΔG values, example (ATP, ADP and AMP: key molecules in biochemical energy transfer) ? (−30.5) − (−31.1) = 0.6 kJ mol−1  Tutorial: G = H − TS 1. ATP → ADP + Pi G° = −30.5 kJ mol−1 (from thermal analysis) H° = −20.1 kJ mol−1 Calculate S° = (all at 37 °C) Comment on the significance of the values of G, H and S 2. H° and S° for the unfolding of a protein are 250.8 kJ mol−1 and 752 J K−1 mol −1 respectively. Above what temperature (in °C) will unfolding of the protein be spontaneous? Comment on the values.  Equilibrium and Energy Equilibrium Energy Defined by equilibrium First law: H constants, Keq Second law: S Define structure of systems G = H – TS Stability when system at Equilibrium when G = 0 equilibrium Spontaneous change when Instability when system not G < 0 at equilibrium Must be a relationship between G and Keq At equilibrium G° = −RTlnKeq  Points about G° = −RTlnKeq True at equilibrium If not at equilibrium, need G − G° = RTlnKeq G is the free energy for the system; G° free energy for the process when at equilibrium Note: penultimate step of derivation gives (Partial) pressure(s) for gases, equivalent to concentrations of components for other phases. Replacement by concentrations, e.g., [C]c, gives equations for Keq G° for the isomerization of glucose-6-phosphate (G6P) to fructose- 6-phosphate (F6P) is 2.1 kJ mol−1. What concentrations of G6P and F6P will be present at equilibrium at 25 °C starting with 0.1 M G6P? 0.0428 − 0.428x = x 0.0428 = 1.428x x = 0.03M = [F6P] [G6P] = 0.1M − 0.03M = 0.07M  Tutorial: G, Keq problem The concentration of a solution of G3P was initially 0.05 M. Isomerase was added. After the mixture came to equilibrium at 25 °C, the concentration of G3P was 0.002 M. Calculate ΔG° for the reaction. What are the implications of the values of ΔG° and Keq for the reaction as given above?  Phases in pharmaceuticals Phases: states of matter, e.g. solid, liquid, gas Phase: homogenous portion of physical material bounded by interfaces A dosage formulation (medicine) will typically contain multiple phases and components Need to understand the factors determining the extent and stability of the differing phases Pharmaceutically, can have gas phases, can have multiple liquid phases and can have multiple solid phases  Phase equilibria Equilibrium between phases can be understood in terms of G = H − TS The phase with the lowest free energy G at temperature T is the most stable at that T For solid phases – enthalpy H is large and −ve (bonding between molecules) – entropy S is small (low disorder) – Hence at low temperatures T, solids tend to be the most stable phase For gas phases – Enthalpy H is close to 0 (no bonding between molecules) – entropy S is large and +ve (high disorder) – gases most stable at high T Liquid phases in between  G as a function of T for a pure substance (i.e. 1 component) Tfus: temperature of G Tfus melting/fusion Tvap Where lines intersect, G = 0 T (free energies of both phases equal) Tvap: temperature of vaporisation  Mixtures of phases and components Say have C components and P phases at equilibrium Each component can be present in each phase Can vary temperature (T) and pressure (P) How much freedom, i.e., flexibility, is there to change parameters without producing new phases or losing phases? The Phase Rule (Gibbs) gives the number of degrees of freedom (F) possible  The Phase Rule F=C−P+2 (“+ 2” allows for variation in temperature and pressure) F: no. of degrees of freedom; C: no. of components; P: no. of phases One component, three phases in equilibrium: F = 0 (triple point) One component, two phases in equilibrium: F=1 Can use Phase Diagrams to illustrate which phases are stable under specific conditions  General phase diagram for a pure substance (i.e., one component) Shows which phase (solid, liquid or vapour) is the most stable at pressure P and temperature T Triple points; three phases at F=C−P+2 Critical point equilibrium C = 1; P = 3; F = 0 Pressure (bar) Water: 0.01 °C, 6.1 mbar Solid Carbon dioxide: −57 °C, 5.1 Liquid bar Critical points: above which substance exists as supercritical Triple Vapour fluid point (combining aspects of liquid and Temperature (ºC) gas phases) For carbon dioxide: 31.1 °C, 73.8 bar  Two component phase diagram To give a 2D plot, data at constant pressure, plot composition vs. temperature Use ‘reduced phase rule’: F = C − P + 1 (fixed pressure P) Temperature (T) Liquid Four regions Two components here All liquid (higher T) called A and B All solid (lower T) Along the ‘x axis’ is Two regions of solid suspended in liquid + solution shown composition in liquid + solid B solid A One has solid A (at higher A%) terms of %A and %B I.e. 100%A/0%B to The other solid B (at higher B%) Solid 0%A/100%B One point (of fixed T and ‘y axis’: temperature (T) composition) at which liquid and 100% A 100% B both solids are in equilibrium: the eutectic point Can only be measured empirically Eutectic point (is characteristic of the system) A system matching the above phase diagram is known as a simple eutectic system Example of a simple eutectic system Example: Naphthalene (C10H8) / Benzene (C6H6) Note: same features as the general diagram (i.e. four regions: all liquid (higher T), all solid (lower T), two regions of solid suspended in solution, one has solid benzene (at higher benzene%), the other solid naphthalene (at higher naphthalene%); one point (ca. −12 °C, 80% C6H6, 20% C10H8) at which liquid and both solid in equilibrium: the eutectic point Note terms ‘Liquidus’ (all liquid above) and ‘Solidus’ (all solid below) Point y: 50%C6H6/50% C10H8 at 60 °C: solution Cool to 10 °C (point z), after an indefinite time, have solid C10H8 suspended in a solution of mostly benzene Ratio of masses of solid and solution given by the lever rule:  Tutorial: Phase diagrams The following thermal analytical data was obtained on mixtures of paracetamol and citric acid (Klimova and Leitner, Thermochim Acta, 2012, 59) Using graph paper, construct a binary phase diagram for the paracetamol/citric acid system Label each region of the diagram and estimate the composition at the eutectic point Note: m.p. paracetamol = 169 ◦C ; m.p. citric acid = 154 ◦C Mole fraction 0.1 0.2 0.3 0.4 0.6 0.8 0.9 paracetamol Solidus temp / ◦C 124 123 122 122 122 123 123 Liquidus temp / ◦ C 151 147 142 137 140 159 164  Mixtures of liquids (A and B) Common in dosage formulation In the pure liquids, A-A and B-B interactions only In the mixture, can also have A-B interactions Number of phases depends on the relative strengths of the A-A/B-B vs. A-B interactions Temperature dependent  Two liquids; upper critical temperature (constant P) 1 liquid phase A-B comparable with A-A, B-B Upper critical temperature Temperature 2 liquid phases A-A and B-B stronger than A-B b c d a 0% A 100% A 100% B 0% B a b: A and B fully miscible, solution of A in B b c: two separate liquid phases; both saturated solutions c d: A and B fully miscible, solution of B in A Examples: hexane/aniline, hexane/nitrobenzene  Two liquids; lower critical temperature (constant P) A-A and B-B stronger than A-B 2 liquid phases Temperature Lower critical temperature 1 liquid phase A-B comparable with A-A, B-B 0% A 100% A 100% B 0% B Can occur when A-B involves complex formation Triethylamine/water, paraldehyde/saline (paraldehyde enema)  Three component systems Can use triangular (ternary) phase diagrams Constant T and P Components 1, 2 and 3 Apices represent pure components Points on sides: 2 components only At point o, composition: 1:2:3 = oa:ob:oc  Alcohol (or surfactant), oil and water systems Alcohol Const. T and P Water/alcohol or oil/alcohol a fully miscible in all proportions Oil/water partially miscible only 1 phase p z w y x b o c Oil q Water 2 phases, one of composition x, the other of composition y xy and wz are tie lines (empirically determined) Upper critical point p  Tutorial Shown below is a ternary phase diagram for an oil/water/surfactant system at a set temperature and pressure. State the composition of the system at points A, B and C. Estimate the composition of the system at point D. State the number of phases present at points E and F. Ternary phase diagrams using triangular graph paper A: 50% solubilizate, 30% water, 20% surfactant B: 60% surfactant, 40% solubilizate, 0% water Line represents dilution of the above with water  Tutorial Data given below for a system of oil, water and alcohol at 25 °C On triangular graph paper, mark in points corresponding to: – Pure oil, pure water, pure alcohol – The five compositions given in the table Sketch in an estimation of the boundary between the region of one liquid phase and the region of two liquid phases One liquid phase Two liquid phases % oil 80 40 10 60 30 % water 10 40 80 30 60 % alcohol 10 20 10 10 10  Multiple solid phases Many pharmaceutical compounds display multiple crystalline solid state forms Known as crystal polymorphism (e.g., paracetamol: at least three polymorphs) Stability can vary with temperature Can also have amorphous (non-crystalline) solid forms The crystal form of the drug product (medicine) has to be controlled Affects solubility and other key properties Phase diagram for compound with two crystal forms Similar to basic one component phase diagram but two solid regions, S1 and S2 These correspond to two crystal polymorphs Which is preferred (more stable) depends on the conditions (temperature and pressure)  Dealing with non-ideality Defined U, H, S, G with reference to ideal gas systems Need to allow for real (non-ideal) systems which are solids or liquids (esp. solutions) For a system, G = f(P, V, T) Say three components, A, B and C. For an ideal system, total G = GA + GB + GC  Chemical potential For an ideal system, total G = GA + GB + GC Not true for a real system For a real system, any change in composition, temperature or pressure depends on the mutual interactions of all components The contribution of each component to the overall free energy is signified by its chemical potential,   Chemical potential () and free energy (G) Real system at constant  and P Say, three components, A, B and C nA moles of A, nB moles of B, nC moles of C Total free energy of the system = GT,P – (i.e., The free energy at a specific temperature and pressure) 𝐺𝑇,𝑃 = 𝜇𝐴 𝑛𝐴 + 𝜇𝐵 𝑛𝐵 + 𝜇𝐶 𝑛𝐶 A = chemical potential of component A B = chemical potential of component B C = chemical potential of component C  Free energy changes in real systems Recall, for a process at equilibrium: G° = −RTlnKeq System not at equilibrium G − G° = RTlnKeq There is a similar equation for an ideal gas: In this, G is the free energy of the gas, P is the pressure of the gas, G° is its free energy under standard conditions and P° is its pressure under standard conditions. n is the number of moles of the gas. is equivalent to a concentration term in an The term equilibrium constant, i.e., [A]a  Free energy changes in real systems For an ideal gas: The equivalent equation for a real system (e.g., a solution) is: 𝜇𝑖 = 𝜇𝑖° + 𝑅𝑇 ln 𝑎𝑖 In which ai = activity of component i It is activities rather than concentration that should strictly be used in expressions of equilibrium constants Concentrations are used as an approximation that is acceptable depending on the usage  Free energy changes in real systems For a real system, such as a solution: 𝜇𝑖 = 𝜇𝑖° + 𝑅𝑇 ln 𝑎𝑖 Activity (ai) is not the same as concentration ([i]) but is directly proportional to it. The proportionality constant is called the activity coefficient () ai = i[i] i = activity coefficient of component i  Free energy changes in real systems Hence for a process such as The free energy can be given as follows (∆𝐺 ° is the free energy under standard conditions): ∆𝐺 = ∆𝐺 ° + 𝑅𝑇 ln 𝐾 And the equilibrium constant is given by: 𝑎𝐶𝑐 𝑎𝐷𝑑 𝐾= 𝑎 𝑏 𝑎𝐴 𝑎𝐵 (depending on the circumstance, the activities can be replaced by concentrations as an acceptable approximation.) Pharmaceutical solutions Acids and bases Ions in solution Assuming solutions in water  Acids and Bases An acid is a proton (H+) donor HA → H+ + A− A base is a proton (H+) acceptor B + H+ → BH+ The above more strictly known as Brønsted acids and bases Lewis acids and bases (a more general definition) – An acid is an electron-pair acceptor – A base is an electron-pair donor  Protons (H+) in water Protons are highly solvated in water Represented by the hydronium ion H3O+ HCl + H2O → H3O+ + Cl− Acid: HCl; base: H2O  Conjugate acids and bases AH + H2O → H3O+ + A− B + H2O → BH+ + HO− AH: acid; B: base A− : conjugate base; BH+: conjugate acid Acids can be neutral, cationic or anionic Bases can be anionic, neutral or cationic Examples  Conjugate acids of bases as acids For any base B, can consider the conjugate acid, BH+, as an acid in its own right E.g: Considering NH4+ as an acid:  General equilibrium for proton transfer Can generally write an equation for the process of an acid transferring a proton to a base (in water) e.g: Equilibrium constant: 𝑎 𝐻3𝑂 + 𝑎 𝐵𝑎𝑠𝑒 𝐾= 𝑎 𝐴𝑐𝑖𝑑 𝑎 𝐻2𝑂  Activity and concentration Equilibrium constants, K, for acid-base equilibria determined by the activities, a, of the components But, provided dealing with dilute solutions… …can replace activities by concentrations as an approximation… …and can assume activity/concentration of water ([H2O]) is constant  Acidity of aqueous solutions Determined by [H3O+] Wide range of concentrations, e.g. 1 mol L−1 to 10−14 mol L−1 Sørensen (1909) defined: pH = −log10 [H3O+] ‘p’ for ‘power’ 10−2 mol L−1 : pH 2; 10−10 mol L−1 : pH 10 Now many functions of the form: pX = −log10 X  Acid dissociation in water Generally considering dilute aqueous solutions [H2O] relatively constant Ka is the acidity constant for acid AH  pKa pKa = −log10Ka Measure of acid strength Strong acid: high Ka, low pKa Weak acid: low Ka, high pKa Compound pKa at 25 °C H2O 15.74 H3O+ −1.74 CH3CO2H 4.76 NH4+ 9.24  Polyprotic acids E.g: H3PO4 [H3O+] in aq. H3PO4 dominated by first dissociation Note: species such as H2PO4- and HPO42- capable of acting as acid or base  Bases in water pKb = −log10Kb  pKa of conjugate acid as a measure of base strength If B a strong base, conjugate acid BH+ a weak acid and vice versa Hence can use pKa of BH+ as a measure of basicity of B  Dissociation of water (at 25 °C) Also known as the autoprotolysis of water Kw is known as the autoprotolysis constant [H3O+][HO−] = 10−14 (−log10[H3O+]) + (−log10[HO−]) = −log10(10−14) pH + pOH = 14 Neutrality at pH 7 At 37 °C, Kw = 13.68, neutrality at pH 6.84  Ka, Kb and Kw Note: AH and A- corresponding acid and conjugate base (or corresponding base and conjugate acid) pKa + pKb = pKw Henderson-Hasselbalch equations Relate pH, pKa and acid/base concentrations  For bases Using pKa of conjugate acid  Buffer solutions Solution of weak acid and its salt (conjugate base) Or a weak base and its salt (conjugate acid) E.g. acetic acid / sodium acetate AcOH weakly dissociated (pKa 4.76); AcONa fully dissociated  Buffering effect Addition of HO- neutralised by AcOH Addition of H3O+ neutralised by AcO- 0.045 mol L-1 soln of AcOH; 0.045 mol L-1 soln of AcONa  Buffering effect E.g. 2.25 × 10−3 mol AcOH and AcONa Add 1.0 × 10−4 mol HNO3 AcO- + H+ → AcOH [AcO-] = (2.25 × 10−3 − 1.0 × 10−4 ) = 2.15 × 10−3 mol [AcOH] = (2.25 × 10−3 + 1.0 × 10−4 ) = 2.35 × 10−3 mol  Buffering effect Buffering effect: pH of solution reduced by addition of acid, but not greatly reduced (from 4.76 to 4.72) 2.25 × 10−3 mol AcOH and AcONa Addition of 1.0 × 10−4 mol HNO3 HNO3 (nitric acid): very strong acid, fully dissociated into H+ and NO3- Concentration of added acid much less than concentration of buffer solution Addition of equal or greater concentrations of added acid: buffering effect would break down.  Common buffers NaH2PO4 / Na2HPO4 pH range 6-8 KH2PO4 / K2HPO4 pH range 6-8 (HOCH2)3CNH2 / (HOCH2)3CNH3+ (TRIS) – pH range 7-9 HEPES pH range 6.8 – 8.2 Worked example: TRIS / TRIS-H+ Buffer pKa TRIS-H+ 8.08 What is the TRIS/TRIS-H+ ratio at pH 8.0 ? 𝐴𝐻+ 𝐴 𝑝𝐾𝑎 = 𝑝𝐻 + log10 𝑝𝐻 − 𝑝𝐾𝑎 = log10 𝐴 𝐴𝐻+ 𝑇𝑅𝐼𝑆 𝑇𝑅𝐼𝑆 −0.08 = log10 = 0.83 𝑇𝑅𝐼𝑆 − 𝐻 + 𝑇𝑅𝐼𝑆 − 𝐻 +  Addition of acid to 0.1 M TRIS/TRIS-H+ 0.1 M = [TRIS] + [TRIS-H+] [TRIS]/[TRIS-H+] = 0.83 at pH 8.0 [TRIS] = 0.83[TRIS-H+]; 0.1 M = 1.83[TRIS-H+] [TRIS-H+] = 0.055 M; [TRIS] = 0.045 M Add 0.005 M H+ (of a strong acid) [TRIS] = 0.040 M; [TRIS-H+] = 0.060 M 𝑇𝑅𝐼𝑆 − 𝐻 + 𝑝𝐻 = 𝑝𝐾𝑎 − log10 𝑇𝑅𝐼𝑆 0.06 = 8.08 − log10 = 7.9 0.04  Addition of acid to 0.02 M TRIS/TRIS-H+ 0.02 M = [TRIS] + [TRIS-H+] [TRIS]/[TRIS-H+] = 0.83 at pH 8.0 [TRIS] = 0.009 M; [TRIS-H+] = 0.011 M Add 0.005 M H+ [TRIS] = 0.004 M; [TRIS-H+] = 0.016 M 𝑇𝑅𝐼𝑆 − 𝐻 + 0.016 𝑝𝐻 = 𝑝𝐾𝑎 − log10 = 8.08 − log10 = 7. 48 𝑇𝑅𝐼𝑆 0.004 Buffer more effective at higher concentration  Tutorial 1. Write an equation for NaHCO3 acting as an acid in water and from that, derive an appropriate Henderson-Hasselbalch equation. 2. The pKa of NaHCO3 is 10.25. How many moles of NaCO3- would be present in a solution of 1.0 moles of NaHCO3 at pH 9.50?  Tutorial Write an equation for the acid dissociation of PhCO2H (benzoic acid) in water and use the equation to derive an appropriate form of the Henderson- Hasselbalch equation. If the pKa of benzoic acid is 4.19, what is the ratio of benzoic acid to benzoate anion, i.e. [PhCO2H]/[ PhCO2−], necessary to provide a solution of pH 3.5?  Ions in solution Behaviour of ions in solutions affected by many factors Degree of ionisation, extent of solvation, ion-ion vs. ion-solvent interactions, external fields, other phenomena In pharmaceutical solutions, need to consider the effect of dissolved ions Need to consider real (rather than ideal) behaviour: i.e. activity rather than concentration  Activity of ions in solution 𝜇𝑖 = 𝜇𝑖° + 𝑅𝑇 ln 𝑎𝑖 ai = activity of component i ai = i[i] i = activity coefficient of component i 𝜇𝑖 = 𝜇𝑖° + 𝑅𝑇 ln 𝑖 + 𝑅𝑇 ln 𝛾𝑖 quantifies deviations from ideality  Activity of ions in solution 𝜇𝑖 = 𝜇𝑖° + 𝑅𝑇 ln 𝑖 + 𝑅𝑇 ln 𝛾𝑖 log10 𝛾± = −𝐴𝑧+ 𝑧− 𝐼 Debye Huckel Limiting Law A is a constant (= 0.509 for water at 25 ◦C) z+ and z- are the charges of the +ve and –ve ions ‘±’ implies mean value for both ions I is the Ionic Strength  Ionic Strength I Depends on the number of cations and anions in solutions Quantifies the ionic field generated by a system of ions in solution 1 𝐼 = ෍ 𝑖 𝑧𝑖2 2  Worked example: Ionic strength and mean activity coefficient of 0.1 M Na3PO4 Na3PO4; fully dissociated in water to 3Na+ and PO43- zNa = 1; [Na] = 3 x 0.1 M zPO4 = 3; [PO4] = 0.1 M I = ½{[(0.3M)12] + [(0.1M)32]} = 0.60 M Log10ϒ± = −(0.509)(1)(3)√(0.6) = − 1.183 ϒ± = 0.066  Mean activity of 0.1 M Na3PO4 a± = ϒ±[±] ([±] = mean ionic concentration of salt) [±] = n√{([+])n+([-])n-} n = n+ + n- For Na3PO4, [±] = 4√{(0.3)3(0.1)} = 0.228 M a± (Na3PO4) = (0.066)(0.228 M) = 0.015 M Significant difference between ionic activity and concentration Activities determine equilibria for ions in solution  Tutorial Determine ionic strength I, mean activity coefficient γ±, mean ionic concentration [±] and mean activity a± for a 0.05 M solution of MgCl2 in water at 25 °C. Assume MgCl2 is fully dissociated in solution.