Pharmaceutics II – Stability

Pharmaceutics II – Stability
Chemical Stability of Pharmaceuticals, Kenneth A Connors, Gordon L Amidon, Lloyd Kennon, John Wiley and Sons, 1979. Chemical Kinetics – The Study of Reaction Rates in Solution, Kenneth A Connors, VCH Publishers, Inc., 1990 STABILITY- Vital component of all product development. All about shelf-life ensures efficacy - physical eg. release characteristics - chemical eg. potency ensures safety - physical eg. tablet weight / dose-dumping - chemical eg. degradants determines storage periods/conditions for formulations

Pharmaceutics II - Stability
TWO AREAS 1 PHYSICAL stability Appearance All formulation types Disintegration Solid DFs Dissolution Solid DFs Hardness Solid DFs Fryability Solid DFs Caking Semi-Solid DFs Cracking Semi-Solid DFs Redispersion Semi-Solid DFs Clarity Liquids + Injectables Particulate Matter Liquids + Injectables Sterility Liquids + Injectables

2 CHEMICAL stability Degradation of API’s All Formulation Types Formation of related substances All Formulation Types Possibly toxic

REGISTRATION AUTHORITIES – MCC of SA Important part of the product registration dossier Require real-time data for all components of Stability Program for determination of shelf-life. Example of Stability Program – following slide Storage conditions - 4°C Depends on Zone - 25°C/60% RH See ICH guidance - 30°C/65% RH (temperate) - 40°C/75% RH (tropical) - 60°C (accelerated)

Stability conducted in final packaging. Any changes in formulation/packaging – re-do stability. Place sufficient packs of product in storage under each condition. Conditions maintained by Controlled Environment Incubators.

HO General Sample Plan of a Stability Study for Tablets Test 0 mths 3 6 12 18 24 36 Limits Appearance U White Assay 90-110 Related Substances <.5% Dissolution 85%<30min Disintegration <15min Hardness XX nm Moisture <0.5% Fryability <1% 25°C/60% RH as above 30°C/65% RH (temperate) as above 40°C/75% RH (tropical) as above 4°C and 60°C (accelerated) abridged

HO Useful Guidelines for Stability Testing MCC of SA - (documents) ICH - International Conference on Harmonisation FDA -

What is Shelf-Life Time taken for any measured parameter to change more than the stated limits allowed Shelf-life for chemical stability: is generally determined by Time taken for the API to reduce from 100% to 90%. Time taken to attain unacceptable levels of toxic degradation products Usually refers to API but can refer to other ingredients. May need to tighten limits for API’s with lower tolerance limits (eg. Critical doses, esp. toxic deg. prod.).

BASIC CONSIDERATIONS Stability test results are unique for a specific formulation. 2. Any formulation change may change API stability e.g. mixing processes, compression etc. 3. Must re-assess stability, perhaps with an abridged stability plan. 4. Temperature changes during manufacture/storage can affect stability – v. imp. 5. Changes may occur during transportation e.g. reach high temperatures in trucks, rail wagon. This must be taken into account for shelf-life and storage conditions. 6. Changes in pH can significantly affect stability. Eg. if no buffers are used.

Main Factors We will consider in detail the two main factors which affect chemical stability: TEMPERATURE - pertinent to all formulations but particularly for solutions. pH – pertinent to liquid aqueous formulations

REACTION KINETICS Order of Reaction Order of Reaction is determined by the molecularity of the reaction. The O-of-R is the sum of the exponents of the reacting species in the rate equation. Eg for D + W → P Rate = –d[D] = k2 [D]1 [W]1 dt O-of-R = 1+1=2 (2nd Order) k2 = 2nd order rate constant Determines the equations which best describe the reaction vs. time profile. Determines the equations to use for calculation and prediction of stability parameters.

3 Useful O-of-Rs Zero Order - independent of concentration First Order - dependent on 1 reacting species. Second Order - dependent on 2 reacting species. In pharmaceutics, degradation reactions generally involve two or more reacting species – first or second order. Units of Rate Constants Second Order - [conc.]-1time-1 First Order - time-1 Zero Order - [conc.]time-1

It is generally assumed that only the concentration of the drug declines with time and that one or more other reacting species such as water or hydrogen ions remains essentially unchanged during the reaction so: First Order - simplifies to Pseudo Zero Order Second Order - simplifies to Pseudo First Order

HO Important Kinetic Parameters t time taken to decline from 100 to 90% i.e. shelf-life k - degradation rate constant t½ - half-life ?

HO HYDROLYSIS in Solution Consider a drug molecule - [D] reacting with - [W] i.e. hydrolysis collision then rearrangement to products if there is sufficient energy. [D] + [W] → [P] eg. Ester hydrolyses to Acid and Alcohol RATE of reaction –d[D] is: dt Rate is proportional to the number of collisions, so Rate is proportional to the concentration of reacting species.

HO As the reaction proceeds [D] and [W] change so Rate or –d[D] α [D][W] - [ ] of two species - Second Order dt Rate or –d[D] = k2 [D][W] k2 = 2nd order rate constant

HO If D is in solution and W is in great excess e.g. 0.1M of D in water where [W] » 55.5M then [D] to [W] ratio is very high (1:555) so any change in [W] due to reaction with D will be minute so [W] can be assumed to be constant and the rate of reaction will not be affected by a changing [W]. Therefore: Rate or -d[D] = k1 [D] [ ] of one species - Pseudo First-Order dt where k1 = k2[W] as [W] is constant k1 = pseudo first-order rate constant

HO If [D] in solution also remains constant (such as in a suspension) then: Rate = k [ ] of no species Pseudo Zero-Order k0 = pseudo zero-order rate constant Where k0 = k1[D] = k2[D][W] Units of Rate Constants Second Order - [conc.]-1time-1 First Order - time-1 Zero Order - [conc.]time-1

HO ACID CATALYSED HYDROLYSIS / DEGRADATION in Solution [D] + [H+] → [P] Second Order In BUFFERED solution where [H+] remains constant the rate of reaction is not affected by a changing [H+]. Rate = k1 [D] - [ ] of one species - Pseudo First-Order where k1 = k2[H+] k1 = pseudo first-order rate constant H+ can be substituted by OH- for alkali degradation. E.g. Aspirin H2O C6H4(OH)COOH + CH3COOH

HO FIRST ORDER CALCULATIONS A typical First-Order reaction can be written as: D → P Therefore the Rate Equation can be written as: -d[D] = k1 [D] Since k1 [D] = k2 [D] [W] dt Integrating the rate equation yields an equation which describes the [D] vs time profile in terms of [D] and t:

HO [D]t i.e. d[D] = k1 dt [D]t at t0 = [D]0 [D]0 i.e. [D]t = [D]0 e-k t or ln [D]t = ln [D]0 – k1 t or Log [D]t = Log[D]0 – k1t 2.303 Note: ln x = Log x Ln 10 = 2.303, Log 10 = 1

HO First-order reaction of a solution [D]0 [D] Time Overall conc. of drug at t=0 All in solution, none in suspension Rate = d[D] = -k1[D] Actual rate of reaction dt changes with time

HO First-order reaction of a solution Linear as Ln[D] vs time Ln [D]90 Ln [D]50 Time Ln[D]t = Ln[D]0 – k1t y = c mx Slope = -k1 (units = time-1) t90 t50

HO Half-Life - t½, t50 t½ is the time it takes for [D]0 to reduce to [D]0/2 i.e. 50% of the initial concentration. t½ can be calculated by substituting into the equation: [D]t = [D]0 x e –k1t [D0 x 0.5] = [D]0 x e –k1t ½ Ln 0.5 = - k1 t½ Ln [D0 x 0.5] = Ln [D]0 – k1t½ e.g If D0 = 1 then t½ = (-) k1 (-)

HO Shelf-Life – t90 t90 is the time it takes for [D]0 to reduce to [D]0 x 0.9 i.e. 90% of the initial concentration or 10% degradation t90 can be calculated by substituting into the equation: [D]t = [D]0 x e –k1t [D0 x 0.9] = [D]0 x e –k1t 0.9 Ln 0.9 = - k1 t0.9 Ln [D0 x 0.9] = Ln [D]0 – k1t0.9 e.g If D0 = 1 then t0.9 = 0.105 k1

APPLICATION Acetyl Salicylic Acid (Aspirin, ASA) has a pH of maximum stability of 2.5. At this pH and at 25ºC the Pseudo First Order rate constant is 5x10-7s-1 Question: What is the t½ and t90 (shelf-life) of ASA? Answer: t½ = = x 106 sec or days 5 x 10-7 t90 = = x 105 sec or days

HO ZERO-ORDER CALCULATIONS The rate expression for zero-order reactions is: -d[D] = k Note – no term for concentration on rhs dt rate is independent of conc. since k0 = k1 [D] = k2 [D] [W] Integration of this equation yields: [D]t i.e d[D] = - k0 dt [D]t at t0 = [D]0 [D]0 i.e [D]t = [D]0 – k0 t Zero-Order Reaction – [D] vs t is Linear

HO Zero-order reaction (suspension) Linear as [D] vs time [D] [D]0 = [D] at t = 100% [D] [D]90 = 90% of [D]0 [D]50 = 50% of [D]0 [D]50 Time [D]t = [D0] – k0t y = c mx Slope = k0 (units = conc.time-1) t90 t50

HO Half-life [D0] x 0.5 = [D0] – k0 t½ t½ = [D0] Zero order k0 Shelf-life [D0] x 0.9 = [D]0 – k0 t90 [D]0 – [D0] x 0.9 = k0 t90 t90 = [D0] Zero order

HO HYDROLYSIS/ACID Degradation in Suspension If [D] in solution also remains constant then: Rate = k [ ] of no species Pseudo Zero-Order where k0 = k2[D][H+] k0 = pseudo zero-order rate constant

HO APPLICATION Ampicillin: pH of maximum stability is 5.8 Rate constant at pH 5.8 is 2 x 10-7s-1 at 35ºC (order?) Drug solubility is 1.1g/100mL Formulated as 125mg/5mL = 2.5g/100mL Question 1: What is the shelf-life of a product formulated as a solution at this pH? Question 2: If this is made as a suspension what is the shelf-life

HO Zero-order reaction of a suspension [D]0 [D]90 [D] [D]sol t t90 Time Soln. Susp. Overall conc. of drug at t=0 Majority of drug in suspension. Solution is saturated with a conc. of [D]sol Degradation occurs at a constant rate of d[D]sol = -k1[D] = -k0 while drug remains in dt suspension i.e. [D] in solution remains constant. d[D] = -k1[D] [D] changes when [D] < [D]sol dt Suspension Solution

Solution: If formulated as a solution e.g. with solubilisers or co-solvents to enhance solubility then: Assume a first-order reaction as all drug is in solution. Therefore t90 = = = 5.3 x 105s. k1 2 x 10-7s-1 = 6.1 days at 35ºC

Suspension: If formulated as a suspension the majority of API is undissolved and in equilibrium with a saturated solution (the formulation vehicle). k1 = x 10-7 s However, assume a zero-order reaction. By definition k0 = k1 [D] Therefore k1 = 2 x 10-7 s-1 x 1.1 g/100mL k0 = 2.2 x 10-7 g/100mL. s-1 Since t90 = 0.1 [D] Zero order k0 t90 = 0.1 x 2.5g/100mL 2.2 x 10-7 g/100mL.s-1 t90 = x 106 s = days

Significant increase in stability when formulated as a suspension. Most paediatric ampicillin products are formulated as dry granules for reconstitution at the time of dispensing and have a shelf-life thereafter of 14 days.

EFFECT OF TEMPERATURE Have looked at the rate equations pertaining to the degradation of an API. Represents degradation reactions occurring under specific conditions e.g. at a specific temperature or pH. k values are specific for a specific set of conditions. What influences k? NB - Temperature and pH To explain the effect of temperature on reaction rates and the associated rate constants we have to look at activation energy theory.

Consider the reaction: A + H+ → B Rate of reaction is prop. to: # of collisions between A and H+ # of collisions with sufficient energy to favour the reaction. Energy required within collisions before the reaction can proceed is determined by the ACTIVATION ENERGY of the reaction. Reacting molecules must attain the energy of the TRANSITION STATE

HO Transition State Theory M‡ ΔG1‡ Influences Rate ΔG1‡ ΔG-1‡ Reactants ΔG0 ΔG0 Influences Extent Products k (k+1 and k-1) A + B P K+1 k (k+1 and k-1) A + B M‡ P K-1

Points to note on Transition State and Activation Energy: A ↔ B 1 A must attain the activated state A++ in order to go to B 2 B can go to A – reverse reaction BUT the activation energy required is greater than going from A to B i.e. the forward reaction is favoured 3 As ΔG0 increases the reaction tends towards B. If the energy of B<<<A then the reaction approaches completion. i.e. the difference in energy between A and B determines the extent of reaction at the end point.

4. The lower the activation energy the greater the number of collisions which will attain the activation complex energy level in a unit time interval. 5. Therefore - the activation energy determines the rate of reaction at a particular temperature i.e. the stability of the API. The SMALLER the activation energy the FASTER the rate of reaction 6. Overall - ↑ Temp → ↑ collisions → ↑ [A++] → ↑ Rate. ↓ Ea → ↑ [A++] → ↑ Rate

7. The relationship between temperature, reaction rate and activation energy is described by the ARRHENIUS EQUATION k = Ae -Ea/RT Where k = reaction rate constant (any order). A = constant Ea = Activation energy of reaction (kcal/mol) T = absolute temperature (K) i.e °C + t °C at which the reaction is taking place. R= Universal gas constant. i.e cal/mol.degree 8.314 x 107 erg/mol.degree Note units

This equation can be re-arranged into a log linear equation which are easy to use: log K = log A Ea Ea . 1 2.303 R T R T y = c mx where 1/T = x This indicates that a plot of log k vs 1/T will be LINEAR with a slope of -Ea 2.303R This is known as a Arrhenius Plot.

HO Arrhenius Plot °C °K /k x 10-3 x 10-3 x 10-3 x 10-3 -1 -2 -3 Log k -4 -5 -6 k373 k353 k333 Slope = Ea 2.303.R k313 1/T (°K) x 10-3

EXPERIMENTAL DETERMINATION OF Ea Conduct stability studies at various temperatures to obtain a number of k’s e.g. using solutions if the k for a hydrolytic degradation is required etc. etc. e.g. at 100°C → k100 (373°K) 80°C → k80 (353°K) 60°C → k60 (333°K) 40°C → k40 (313°K) Use elevated temperatures to speed up the reaction and reduce experimental time when reactions are slow, which is the case for most pharmaceuticals.

Determine k at various elevated temperatures [D] 313°K 333°K 353°K 373°K Time

HO Determine k at various elevated temperatures Ln[D] Or Log [D] (1/2.303) Time k313 k333 k353 k373

Plot kT vs 1/T as per the arrhenius plot. Determine slope of plot. Determine Ea since Slope = Ea/R using Ln = Ea/(R.2.303) using Log Can use: linear paper – plot ln or log Ln-linear paper – remember that ln= 2.303log.

HO Arrhenius Plot °C °K /k x 10-3 x 10-3 x 10-3 x 10-3 -1 -2 -3 Log k -4 -5 -6 k373 k353 k333 Slope = Ea 2.303.R k313 1/T (°K) x 10-3

From an Arrhenius Plot, the Activation Energy of a reaction can be calculated. Once the Activation Energy is known, the reaction rate constant k can be calculated for any temperature. t90 and t½ can then be calculated at any temperature. Note: Activation Energy of a reaction is specific for a reaction conducted under specific conditions. However, it does not change with temperature.

The Arrhenius Equation can be re-arranged into several useful forms: log ( k2 ) = Ea ( ) ( k1 ) R ( T2 T1 ) log of the ratio of k2 / k1 or log k2 = Ea (T2 – T1) k R.T2.T1 Where k1 and k2 are the rate constants at temperature T1 and T2 in °K (°C+273).

HO Ea’s for some Common Pharmaceutical Degradation Reactions. Compound Reaction Ea (kcal/mol) Ascorbic Acid Aspirin Atropine Benzocaine Chloramphenical Epinephrine Procaine Thiamine Oxidation Hydrolysis 23 14 19 20 Range of Ea for pharmaceuticals is generally between kcal/mol

Worked Example – Sulphacetamide Antibacterial agent used as the Na+ salt in ophthalmic solutions. Ea = 22.9 kcal/mole at pH 7.4 Degradation is pH independent between pH 5 and 11. k1 at 120° = 9.0 x 10-6 sec-1 (temp. for sterilisation). In the pH range of 5-11, sulphacetamide exists as an anion with a pKa of 5.21 (acidic). The ionised form is more stable than the unionised form. Degradation is by hydrolysis – i.e. addition of water, which is catalysed by H+ and OH- at the extremes of pH.

HO

k at °C = 9 x S-1 1st order 120°C = 393°K = T1 25°C = 298°K = T2 k at 25° = k25 Question – What is the shelf-life of this opthalmic solution at 25°C Question – What percentage will be degraded during the sterilisation process at 120°C for 30 minutes.

1. Calculate k2 ie k at 25°C log k25 = Ea ( ) k R (T2 T1 ) log k25 = ( ) k x ( ) = ve / +ve NB Therefore k25 = x (antilog of -4.06) k120 Therefore k25 = 8.7 x x 9 x S-1 = 7.85 x S-1

Now t90 = First Order k25 Therefore t90 = S-1 7.85 x 10-10 t90 = 1.34 x 108 S t90 = approx years. This agrees with the monograph shelf life of 4.2 years at 25°C

What percentage would degrade during the sterilisation process at 120°C for 30 mins? Will it be significant? k at °C = x S-1 first order t = 30min = 1800 s Ln [D]t = Ln [D]0 – k1 t Ln [D]t - Ln [D]0 = – x 30 x 60 units NB Ln (D/D0) = ve NB D/D0 = Therefore % degraded = 1.62%

HO Q10 values and their use in the estimation of shelf-life A Q10 value is the ratio of reaction rate constants (k) for two reactions 10°C apart. i.e. This ratio is essentially constant for a particular reaction irrespective of which 10°C interval is considered. However, the Q10 value depends on the activation energy of the reaction. Ln[D] Time 310° 300°

HO If the activation energy is high then the Q10 value is high and visa verse. Approximate Q10 values for a range of Ea values are as follows: Ea Kcal/mol Q10 ΔT Δk 14 2 +10°C X2 Doubles 19 3 X3 Triples 24 4 X4 Quadruples 24 14 19 Ln[D] Time 310° 300° Ln[D] Time 310° 300° Ln[D] Time 310° 300°

HO 40°C 313°K 30°C 303°K Q10 = k40 = 2 k30 Q10 = k40 = 3 k30 Q10 = k40 = 4 k30 Ea = 14 Kcal/mole Ea = 19 Kcal/mole Ea = 24 Kcal/mole 1/T (°K) x 10-3

HO If temperature change is more than +10°C then: If ΔT = 25°C ΔT = = Therefore

HO If storage temperature is increased above the stated storage temperature then: Ea Kcal/mol Q10 ΔT ΔT / 10 QΔT Δk t90 14 2 +25 2.5 5.7 (22.5) x 5.7 1/5.7 19 3 15.6 (32.5) x 15.6 1/15.6 24 4 32 (42.5) x 32 1/32

HO If storage temperature is decreased below the stated storage temperature then: ΔT = -25°C ΔT = = Therefore Ea Kcal/mol Q10 ΔT ΔT / 10 QΔT Δk t90 14 2 -25 -2.5 1/5.7 (2-2.5) x 0.175 x 5.7 19 3 1/15.6 x 0.064 x 15.6 24 4 1/32 x 0.031 x 32

HO However, in the Pharmacy you don’t generally know k but you do know t90 (shelf-life). You know that there is an indirect relationship between k and t90 therefore k can be substituted for 1/t90 so:

HO Q10 values can be used to calculate very approximate changes in shelf-life when products are stored at higher or lower temperatures than stipulated. When calculating a new shelf-life the safest possible estimate with respect to patient safety should be used. Considering that pharmaceutical degradation reactions are generally in the range of 13 to 24 kcal/mol with respective Q10 values of 2 to 4: A Q10 of 4 should be used to calculate a decrease in the shelf-life after storage at elevated temperature as this will generally over-estimate the rate of degradation and under-estimate the new shelf-life. A Q10 of 2 should be used to calculate an increase in the shelf-life after storage at reduced temperature as this will generally over-estimate the rate of degradation and under-estimate the new shelf-life.

HO 1 The expiration period of a reconstituted product is 18 hours at room temperature. Estimate the expiration period when the product is stored in the fridge. 2 A newly reconstituted product is labeled to be stable for 24 hours in the fridge. What is the estimated shelf-life at room temperature. 3 A product has an expiry date of 1 year when stored in the refrigerator. The product has been stored for one month at room temperature. If the product is returned to the refrigerator what is it=s new expiry date. 4 A reconstituted suspension of Ampicillin is stable for 14 days in the refrigerator. If the product is left out of the fridge for 12 hours what is the reduction in the expiry date. 5 If we know that the Ea of ampicillin at pH=5 is 18.3 kcal/mol then calculate the actual expiry date reduction.

The expiration period of a reconstituted product is 18 hours at room temperature. Estimate the expiration period when the product is stored in the fridge. hrs at 25°C X hrs at 5°C ΔT = -20°C Therefore if Q10 = 2 then If Q10 = 4 then so since so since Therefore X = 72 hours Therefore X = 288 hours Use 72 hours as a conservative estimate (decrease temp. – use 2)

2 A newly reconstituted product is labeled to be stable for 24 hours in the fridge. What is the estimated shelf-life at room temperature hrs at 5°C 0 X hrs at 25°C ΔT = 20°C Therefore if Q10 = 2 then If Q10 = 4 then so since so since Therefore X = 6 hours Therefore X = 1.5 hours Use 1.5 hours as a conservative estimate (increase temp. – use 4)

3 A product has an expiry date of 1 year when stored in the refrigerator. The product has been stored for one month at room temperature. If the product is returned to the refrigerator what is its new expiry date months at 5°C 0 1 month at 25°C 0 equivalent to X months at 5°C ΔT = 20°C (25°C to 5°C) Therefore if Q10 = 2 then If Q10 = 4 then so since so since Therefore X = 4 months Therefore X = 16 months Use 16 months as a conservative estimate so the product should be discarded.

4 A reconstituted suspension of Ampicillin is stable for 14 days in the refrigerator. If the product is left out of the fridge for 12 hours what is the reduction in the expiry date days at 5°C 0 12 hours at 25°C 0 equivalent to X months at 5°C ΔT = -20°C (25°c TO 5°c) Therefore if Q10 = 2 then If Q10 = 4 then so since so since Therefore X = 2 days Therefore X = 8 days Use 8 days as a conservative estimate so the product can be stored at 5°C for a further 6 days.

The Effect of Catalysis on Reaction Rate Does a Catalyst - Effect rate? Effect equilibrium? Get transformed? Get consumed? Get regenerated? Catalysis and transition state theory - Explained by effect on activation energy

HO Transition State Theory M‡ uncatalysed. ΔG‡ uncat. M‡ catalysed. ΔG‡ Influences Rate ΔG‡ cat. ΔG-1‡ Reactants ΔG0 ΔG0 Influences Extent Products

Types of Catalysis in Pharmaceutics Specific Acid Catalysis Catalyst is the solvated proton from an acid i.e. H+ which is H3O+ in solution. Eg HCl H+ + Cl- then H+ + H2O H3O+ Specific Base Catalysis Catalyst is the solvated hydroxyl ion from a base i.e. OH- in solution. Eg NaOH Na+ + OH- General Acid Catalysis Catalyst is a proton other than an hydronium ion. Any “Bronstead Acid” or compound which can donate a proton eg. NH4 + General Base Catalysis Catalyst is a proton acceptor other than OH- such as a “Bronstead Base” which acts as a proton acceptor by sharing an electron pair.

Value of studying catalysis - ascertain effect of catalyst on reaction rate - ascertain what compounds in a formulation act as catalysts - can recognise potential catalysts in proposed formulations - can formulate to minimise a catalytic effect on stability.

The Effect of pH on Reaction Rate and H+ and OH- as catalysts pH is a major factor affecting reaction rates in solution as reactions are often catalysed by H+ and/or OH- ions Other factors such as temperature ionic strength of the solution solvent composition and additives The effect of pH on reaction rate can be determined by determining the degradation profiles and first-order rate constants of reactions (solutions) at various pHs. If the log of the rate constants obtained at various pHs are plotted against pH, a pH rate profile is obtained

HO pH rate profiles are generally one or a combination of three basic shapes: 1 V-graph 2 Sigmoidal 3 Bell-Shaped Log K pH Log K pH Log K pH

HO V-Graphs Obtained for specific acid and specific base catalysis of NON-IONISABLE compounds in aqueous solution . OR of ionisable compounds where ionisation does not affect stability. Consider the reaction of D Products e.g. The hydrolysis of an ester with water Three possible reaction pathways co-exist : 1 Catalysis by H+ - reaction proceeds under specific acid catalysis 2 Catalysis by 0H- - reaction proceeds under specific base catalysis 3 Un-catalysed – reaction proceeds without catalysis. i.e. Hydrolysis by water without any influence from H+ or OH- ions. Overall rate of reaction at a particular pH will be the sum of the individual rates of reaction for the three possibilities above.

HO Hypothetical Rate Equation is: D H Un-C OH- P P P rate = k1 [D][H+]n k2 [D] k3 [D][0H-]m acid catalysed un-catalysed base catlaysed Since this is hydrolysis, H2O is also involved in the reaction but kn takes this into account as the concentration of H2O remains constant. Remember Rate or –d[D] = k2 [D][H+] 2nd order dt = k1 [D] 1st order

HO The overall rate equation will be: rate = k [A] Therefore k [D] = k1 [D][H+]n k2 [D] k3 [D][0H-]m k = k1 [H+]n k k3 [0H-]m Or k = k1 [H+]n k k3 . Kw m [H+] Since Kw = [H+][OH-] = 1.0 x at 25°C (ion product of water) n and m are the orders of reaction with respect to H+ and OH- and are = 1 k1, k2 and k3 are the rate constants of the specific acid, base or uncatalysed reactions.

HO Now Consider if the pH of the reacting solution is low: H+ is high so only first term is significant the second and third terms can be ignored Therefore k = k1 [H+]n Taking Logs and rearranging for H+ log k = log k1 – n.pH Therefore a plot of log k vs pH will be a straight line with a gradient of n. For specific acid catalysis n = 1 therefore the gradient will be -1.

HO Now Consider if the pH of the reacting solution is high: OH- is high so only third term is significant the first and second terms can be ignored Therefore k = k3 [0H-]m Taking Logs and rearranging for H+ log k = log k3 – m.p0H log k = log k3 + m.pH -14 Therefore a plot of log k vs pH will be a straight line with a gradient of +m. For specific base catalysis m = 1 therefore the gradient will be +1.

If degradation studies are conducted at high and low pHs then the complete profile can be constructed and k2 determined. When all three ks are known the overall k can be accurately calculated for any pH. k = k1 [H+]n k k3 [OH]m Any pH – rate profile is only valid for one temperature. Calculation of rates at temperatures different from which the pH-rate profile was determined will require the use of the Arrhenius equation as you have already seen.

Log k3 Log k1 n (-1) m (+1) Log k (overall) pH of max. stability Log k2 1 pH Pharmaceutics II - Stability HO Straight lines with gradients of -1 and 1 demonstrate direct relationship between pH or pOH and k(overall) and specific acid or base catalysis respectively. Position of the inflection point is dependent on the relative magnitudes of k1 and k3 .

Usually k1 << k3 therefore the inflection point is usually on the acidic side of neutral. If k1 >> k3 then the inflection point will be on the alkali side of neutral. For specific acid and specific base catalysis, maximum stability occurs at the inflection point when: k1[H+]n = k3[OH-]m k (overall) at the point of inflection is approximately equal to k2 , however, k2 actually resides below the inflection point, the extent of which depends on the magnitudes of k2 relative to the other two ks.

Sigmoidal Graphs This shape of pH-rate profile is usually obtained for weak acids and weak bases (ionisable) where degradation is not significantly catalysed by H+ or OH- but ionisation plays a major role in the uncatalysed degradation of the molecule. Consider the degradation of an ionisable weak acid: HD H D- k† k‡ P P k† and k‡ are rate constants where the degradation rates of the two different species is different. Catalysis by H+ or OH- will complicate things considerably.

Hypothetical rate equation is rate = k†[HD] k‡[D-] Overall rate = k[Dtotal] Therefore k[Dtotal] = k†[HD] + k‡[D-] (1) However, ka = [H+][D-] (2) [HD] Ratio of D- to HD (degree of ionisation) depends on pH After combining and rearranging (1) and (2), k = k†[H+] + k‡ ka [H+] + ka This describes the rate of reaction in terms of H+ concentration (pH) and the dissociation constant ka of the compound.

HO Considering again the degradation of an ionisable weak acid: HD H D- k† k‡ P P It follows that the pH-rate profile for the degradation of this week acid will follow the ionisation profile since the ratio between HD and D- will ultimately determine the overall degradation rate.

HO Ionisation Profile HD H+ + D- 100% % HD D- % HD %D- 0% pKa % 1 pH

HO HD H+ + D- k† k‡ P P If k† >> k‡ k† Log k Log k pKa 1 pH Max HD Max D- Min D Min HD

HO HD H+ + D- k† k‡ P P If k† > k‡ k† Log k Log k pKa 1 pH Max HD Max D- Min D Min HD

HO HD H+ + D- k† k‡ P P If k† = k‡ k† Log k Log k pKa 1 pH Max HD Max D- Min D Min HD

HO HD H+ + D- k† k‡ P P If k† < k‡ k‡ Log k Log k pKa 1 pH Max HD Max D- Min D Min HD

HO HD H+ + D- k† k‡ P P If k† << k‡ k‡ Log k Log k pKa 1 pH Max HD Max D- Min D Min HD

HO Considering again the degradation of an ionisable weak base: BH B + H+ k† k‡ P P It follows that the pH-rate profile for the degradation of this week base will follow the ionisation profile since the ratio between B and BH+ will ultimately determine the overall degradation rate.

HO Ionisation Profile BH+ B + H+ 100% % BH B % BH % B 0% pKa % 1 pH

HO BH+ B + H+ k† k‡ P P If k† >> k‡ k† Log k Log k pKa 1 pH Max BH Max B Min B Min BH+

HO BH+ B + H+ k† k‡ P P If k† << k‡ k‡ Log k Log k pKa 1 pH Max BH Max B Min B Min BH+

HO Bell Shape - 1 This shape of pH-rate profile can be obtained for weak acids or bases (ionisable) which are di-acidic or di-basic (release or accept two protons) i.e. the compound has two pKa’s Where degradation is not significantly catalysed by H+ or OH- but ionisation plays a major role in the uncatalysed degradation of molecule. Consider the degradation of an ionisable weak acid: H2D H HD H+ + D- - k† k‡ k‡ † P P P k†, k‡ and k‡ † are rate constants where the degradation rate of the three different species is different.

HO Ionisation Profile H2D H+ + D- 2H+ + D- - 100% % pKa % Ionisation 0% % 1 pH

HO H2D H+ + D- 2H+ + D- - k† k‡ k‡ † P P P If k† << k‡ >> k‡ † Log k Log k 1 pH Max H2D Max HD Max D- -

Practically, pH-Rate profiles are useful as they: 1. Provide information on rate vs pH (obvious). 2. Provide info on the degradation process – specific acid/base catalysis or general acid/base catalysis. 3. Provide info on how to limit degradation 4. Can be determined by measuring the concentration vs time under different conditions over the pH range if the rate equation is complex. 5. Can be calculated from minimal experimental data if the rate equation is known.

Eg. 1 Determine at what pH AMPICILLIN in solution is most stable. Method: 1. Determine the degradation rate at high and low pHs where degradation is fast. 2. Derive the rate equation from knowledge of the reaction. 3. Calculate points on the pH-rate profile and plot Log k vs pH. 4. May need to determine pH-rate profile at different temperatures. 5. Can calculate Activation Energy at any temperature and therefore k and t90 at any temperature. .

55°C Does activation energy change with pH? 45°C 35°C -4 Log k Log k sec-1 pH of Max Stability -6.6 1 pH

Eg. 2 The pH=rate profile of Aspirin is presented below. Is this product suitable for reconstitution with water before dispensing. - Unbuffered or buffered at what pH? - Estimate the t90 of an Aspirin solution at the pH of maximum stability and buffered at pH 7.0 where pH change with time would not be significant. Log k Log k sec sec-1 pH

Log k at pH 3.5 = therefore k = 3.00 x s-1 Log k at pH 7.0 = therefore k = 3.10 x 10-6 s-1 At pH 3.5 t90 = 0.105/3.00x10-7 = 3.5x105 s = approx. 4 days At pH 7.0 t90 = 0.105/3.10x10-6 = 3.3x104s = approx. 9 hours For a reconstituted product, only reconstitution at the time of dosing is really feasible. Otherwise, alternative methods of stabilising aspirin solution must be employed.

This profile represents: 1. pH 1 – 3.0 Specific acid catalysis (H+) 2. pH 3.0 – 11 Sigmoid due to ionisation with plateau where ionisation is complete but without OH- catalysis 4 pH 11 – 14 Specific base catalysis (OH-) -4 -5 Log k Log k sec-1 -6 -7 pH

Assuming RC00H represents aspirin then 1. RCOOH Products Specific acid catalysis 2. RCOOH Products 3. RCOO- Products 4. RCOO- Products Specific base catalysis 1 and 4 represent the V-shaped portion of the profile where [H+] and [OH-] are in high concentration respectively. 2 and 3 represent the sigmoid portion of the profile where the less stable ionised form which has decreased stability increases in concentration. k1 H+ kI Not catalysed significantly by H+ or OH- but ionisation degreases stability due to increased ionised moiety H2O kII H2O k3 OH-

The rate equation which accounts for this profile can then be written as: koverall [RCOOH] = k1[RCOOH][H+] + kI[RCOOH] + kII[RCOO-] + k3[RCOO-][OH-] From the profile it is evident that k3 > k and kII > kI

Eg. 2 What will an acceptable pH be for a solution of pilocarpine (eye drops) to have a t90 of 4 months if stored at 25°C. Want a t90 = 4 months = seconds Since t90 = 0.105/k k = 1x10-8 s-1 Log k = -8 acceptable range = 4.5 – 7.0 -2 -4 Log k Log k sec-1 -6 -8 -10 pH

Pharmaceutics II – Stability

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