Pharmaceutics Solutions Week 2 PDF

Solutions PY4030/PY5130 Dr Gianpiero Calabrese Room MB1037, [email protected] Learning Outcomes Define the terms: solution, solubility and solubility constant, and their importance Appreciate the consequences of adding a common ion to an ionic solution Define the terms pKw, Ka, Kb, pH an...

Solutions PY4030/PY5130 Dr Gianpiero Calabrese Room MB1037, [email protected] Learning Outcomes Define the terms: solution, solubility and solubility constant, and their importance Appreciate the consequences of adding a common ion to an ionic solution Define the terms pKw, Ka, Kb, pH and pOH Use the Henderson-Hasselbalch equation to calculate the extent of ionization of a weak acid or base at different pH values Define buffers and appreciate their importance What are solutions? Homogeneous mixtures of two or more pure substances In a solution, the solute is dispersed uniformly throughout the solvent Why are solution important? In order to be absorbed and thus exert an action a drug must be in solution Swallow a tablet → Tablet Disintegrates & Dissolves → Drug is absorbed Drug Tablet (SOLID) Dissolution Drug in solution in GI tract Absorption Drug in solution in Blood Water And Drugs 50-65% of our body mass is water All reactions occurring in the body do so in aqueous solutions Therefore we need to understand water as a medium & how drugs behave in aqueous solutions Many drugs (APIs) are weak acids or bases: • They dissociate partially into ions when dissolved in water • Ionized & unionized drugs behave differently -E.g. absorption -Making different salt forms is common Solvation a.k.a. Dissolution How does a solid dissolve into a liquid? What ‘drives’ the dissolution process? What are the energetics of dissolution? How Does a Solution Form? 1. Solvent molecules attracted to surface ions. 2. Each ion is surrounded by solvent molecules. 3. Enthalpy changes (∆H) with each interaction broken or formed. Ionic solid dissolving in water Example The ions are solvated (surrounded by solvent). If the solvent is water, the ions are hydrated. The intermolecular force here is ion-dipole. Energy Changes in Solution To determine the enthalpy change, we divide the process into 3 steps. 1. Separation of solute particles. 2. Separation of solvent particles to make ‘holes’. 3. Formation of new interactions between solute and solvent. Start End Start End Enthalpy changes DHsoln = DH1 + DH2 + DH3 The enthalpy of solution, DHsoln, can be either positive or negative. Enthalpy is only one side of the story (entropy). DHsoln (MgSO4)= -91.2 kJ/mol --> exothermic DHsoln (NH4NO3)= 26.4 kJ/mol --> endothermic Degree of saturation Unsaturated solution -Less than the maximum amount of solute for that temperature is dissolved. -No solid remains in flask. Degree of saturation Saturated solution -Solvent holds as much solute as is possible at that temperature. -Undissolved solid remains in beaker. -Dissolved solute is in dynamic equilibrium with solid solute particles. Solubility -Solubility is an equilibrium term -Between solid and saturated solution -It is dynamic Solid Solution For dissolution to happen -Chemists use the axiom “like dissolves like”: ➢Polar likes polar! For dissolution to happen -Chemists use the axiom “like dissolves like”: ➢Polar substances tend to dissolve in polar solvents. ➢Nonpolar substances tend to dissolve in nonpolar solvents. -Attractions between solute particles must be overcome by attractions of the solvent for the solute particles -The stronger the intermolecular attractions between solute and solvent, the more likely the solute will dissolve. Example The Solubility Constant Consider the equilibrium Ca(OH)2(s) Ca2+(aq) + 2 OHThe equilibrium constant is called the solubility constant, Ks Ks = [Ca2+] [OH-]2 Square brackets indicate concentration Temperature -Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature. -The opposite is true of gases. Higher temperature drives gases out of solution. Common Ion Effect Since Ks is constant (for Ca(OH)2 Ks = 5.5 x 10-6 mol3 L-3) If we add OH- ions (eg. by adding NaOH) [Ca2+] must decrease to keep Ks constant# Ca(OH)2(s) Ca2+(aq) + 2 OH- Common Ion Effect [Ca2+] must decrease to keep Ks constant (Le Chatelier's principle) Ca(OH)2(s) Ca2+(aq) + 2 OH- Dissociation of Water • Water disassociates into ions • H2O(l) = H+(aq) + OH-(aq) Strictly speaking the hydroxonium is produced H2O(l) + H+(aq) = H3O+(aq) • In pure water at 25 °C [H+] = [OH-] = 10-7 mol dm-3 • The autoprotolysis constant of water is Kw • Kw = [H+][OH-] mol2dm-6 = 10-14 Dissociation of Water • • • [H+] may be altered by adding acid or base to the water [H+] varies over a wide range: 10 to 10-15 mol dm-3 We use a log scale: pH = - log10 ([H+]/mol dm-3) pOH = - log10 ([OH- ]/mol dm-3) • For pure water at 25 °C pH = 7.0 10x is the inverse function of log10 pKw = pH + pOH = 14 • pH 7 is neutral, but should not be considered ‘normal’ o Skin typically pH 4-5.5 o Thames Water typically pH 7-8 Lowery-Brønsted Theory An acid is a substance which can donate a proton A base is a substance which can accept a proton pH = -log10[H+] pH ranges from 1 – 14 Dissociation of Acids Take a weak acid HA, in water its dissociation is given by: HA H + + A- Extent of ionisation governs the ability of the acid to donate a proton & determines the strength of the acid Equilibrium constant for this equilibrium is the dissociation constant Ka  H 3O + A−  Ka = HA Taking logs and rearranging we have pK a  HA = pH + log A−  (Henderson – Hasselbalch eqn) Dissociation of Bases • We can write the ionisation of a base (B) as BH+(aq) + H2O B(aq) + H3O+(aq) • And  BH +  pK a = pH + log B  • Alternatively, in water a base picks up a proton; B(aq) + H2O BH+(aq) + OH-(aq) • Where the equilibrium constant Kb is  BH OH  = + Kb B  − pKb = -log10 Kb / mol dm-3 • pKb of a base B is related to the pKa of the conjugate acid BH+ via • pKa + pKb = pKw = 14 Buffer Solutions A buffer solution is a solution formed by a weak acid (or a weak base) and a salt of the conjugate base (or conjugate acid) and is able to withstand small additions of acid or base without much change in the pH of the solution. One example is a mixture of ammonia NH3 and ammonium chloride NH4Cl For buffers:  Salt pH = pKa + log Acid  pH, pKa and ionisation • pH - pKa = log10([A-]/[HA]) • 10(pH – pKa) = [A-]/[HA] • Aspirin, pKa = 3.5; Plasma pH is 7.4 7.4 - 3.5 = 3.9 103.9 = 7943 Ratio of [A-]:[HA] = 7943:1 • pH of stomach is 2 (approx) 2 - 3.5 = -1.5 10-1.5 = 0.0316 Ratio of [A-]:[HA] = 0.0316:1 References And Further Reading • Aulton's pharmaceutics : the design and manufacture of medicines, Sixth edition / edited by Kevin M.G. Taylor, Michael Aulton. • Physicochemical Principles of Pharmacy. A. Florence & D. Attwood, 5th edition, Pharmaceutical Press (2011).