Pharmaceutics II – Stability STABILITY-Vital component of all product development. All about...

Pharmaceutics II – Stability

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 1

1. Chemical Stability of Pharmaceuticals, Kenneth A Connors, Gordon L Amidon, Lloyd Kennon, John Wiley and Sons, 1979.

2. Chemical Kinetics – The Study of Reaction Rates in Solution, Kenneth A Connors, VCH Publishers, Inc., 1990

Pharmaceutics II - Stability

TWO AREAS 1 PHYSICAL stabilityAppearance All formulation typesDisintegration Solid DFsDissolution Solid DFsHardness Solid DFsFryability Solid DFsCaking Semi-Solid DFsCracking Semi-Solid DFs

Redispersion Semi-Solid DFsClarity Liquids + InjectablesParticulate Matter Liquids + InjectablesSterility Liquids + Injectables

2

2 CHEMICAL stability

Degradation of API’s All Formulation Types

Formation of related substances All Formulation Types

Possibly toxic


3

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°CDepends on Zone - 25°C/60% RHSee ICH guidance - 30°C/65% RH (temperate)

- 40°C/75% RH (tropical)- 60°C (accelerated)

4


5

• 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.


Test 0 mths 3 6 12 18 24 36 Limits

Appearance U U U U U U U White

Assay U U U U U U U 90-110

RelatedSubstances

U U U U U <.5%

Dissolution U U U U U U U 85%<30min

Disintegration U U U U U <15min

Hardness U U U U U U U XX nm

Moisture U U U U U <0.5%

Fryability U U U U U <1%

6

Pharmaceutics II - StabilityGeneral Sample Plan of a Stability Study for Tablets

25°C/60% RH as above30°C/65% RH (temperate) as above40°C/75% RH (tropical) as above

4°C and 60°C (accelerated) abridged

HO


Useful Guidelines for Stability Testing

• MCC of SA - www.mccza.com (documents)

• ICH - International Conference on Harmonisationwww.ich.org

• FDA - www.fda.gov

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http://www.mccza.com/

http://www.ich.org/

http://www.fda.gov/

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.).

8


BASIC CONSIDERATIONS

1. 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. 9


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

10


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.11


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 12


• 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

13



Important Kinetic Parameters

• t90 - time taken to decline from 100 to 90%

i.e. shelf-life

• k - degradation rate constant

• t½ - half-life ?

14

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. 15

Pharmaceutics II - StabilityHO

• As the reaction proceeds [D] and [W] change so

• Rate or –d[D] α [D][W] - [ ] of two species - Second Orderdt

• Rate or –d[D] = k2 [D][W] k2 = 2nd order rate constant

dt

16


• 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 17


• If [D] in solution also remains constant (such as in a suspension) then:

• Rate = k0 - [ ] 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

18



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

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• 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:

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[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 = 2.303 Log x» Ln 10 = 2.303, Log 10 = 1

21

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• [D]0

[D]

Time


22

Rate = d[D] = -k1[D] Actual rate of reaction dt changes with time

Overall conc. of drug at t=0All in solution, none in suspension

First-order reaction of a solution

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• Ln [D]90

Ln [D]50

Time


23

Slope = -k1 (units = time-1)

Ln[D]t = Ln[D]0 – k1t y = c - mx

First-order reaction of a solutionLinear as Ln[D] vs time

t90 t50

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• 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 -0.693 0

• t½ = 0.693 (-) k1 (-)

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• 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 -0.105 0

• t0.9 = 0.105

k1

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26

• 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 = 0.693 = 1.39 x 106 sec or 16 days5 x 10-7

• t90 = 0.105 = 2.1 x 105 secor 2 days

5 x 10-7


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• ZERO-ORDER CALCULATIONS

• The rate expression for zero-order reactions is:

• -d[D] = k0 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

[D]0 [D]0 = [D] at t = 100%

[D]90 [D]90 = 90% of [D]0

[D]50 = 50% of [D]0

[D]50

Time


28

Slope = k0 (units = conc.time-1)

[D]t = [D0] – k0t y = c - mx

Zero-order reaction (suspension)Linear as [D] vs time

t90 t50

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• Half-life [D0] x 0.5 = [D0] – k0 t½

• t½ = 0.5 [D0] Zero order

k0

• Shelf-life [D0] x 0.9 = [D]0 – k0 t90

• [D]0 – [D0] x 0.9 = k0 t90

• t90 = 0.1 [D0] Zero order

k0

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HYDROLYSIS/ACID Degradation in Suspension

• If [D] in solution also remains constant then: • Rate = k0 - [ ] of no species Pseudo Zero-Order

• where k0 = k2[D][H+]

• k0 = pseudo zero-order rate constant

HO


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• APPLICATION

• Ampicillin: pH of maximum stability is 5.8Rate constant at pH 5.8 is 2 x 10-7s-1 at 35ºC (order?)Drug solubility is 1.1g/100mLFormulated 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

[D]0

[D]90

[D] [D]sol

t90 t90 Time

Soln. Susp.


32

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.

Overall conc. of drug at t=0Majority of drug in suspension.Solution is saturated with a conc. of [D]sol

Zero-order reaction of a suspension

d[D] = -k1[D] [D] changes when [D] < [D]sol dt

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Suspension

Solution


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• 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 = 0.105 = 0.105 = 5.3 x 105s.

k1 2 x 10-7s-1

= 6.1 days at 35ºC


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• Suspension: If formulated as a suspension the majority of API is undissolved and in equilibrium with a saturated solution (the formulation vehicle).

• k1 = 2 x 10-7 s-1 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 = 1.136 x 106 s = 13 days


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• 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.


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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.


37

• 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

•


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Transition State Theory

ΔG1‡

Influences Rate

ΔG-1‡

ΔG0

ΔG1‡

ΔG0

Influences Extent

P

P

A + B

A + B

k (k+1 and k-1)

K+1

K-1

M‡

M‡

Reactants

Products

HO

k (k+1 and k-1)


39

• 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 energyrequired 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.


40

• 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


41

• 7. The relationship between temperature, reaction rate andactivation 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. 273.16 °C + t °C at which the reaction is

taking place.• R= Universal gas constant.• i.e. 1.987 cal/mol.degree• 8.314 x 107 erg/mol.degree

Note units


42

• 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 2.303R 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.


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• Arrhenius Plot

• -1

-2

-3Log k -4

-5

-6 1 2 3 4 5 6

1/T (°K) x 10-3

Slope = -Ea 2.303.R

°C °K 1/k40 313 3.19 x 10-3 60 333 3.00 x 10-3 80 353 2.83 x 10-3 100 373 2.68 x 10-3

k373

k353

k333

k313

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44

• 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.


45

• Determine k at various elevated temperatures

313°K

333°K353°K373°K

[D]

Time


46

• Determine k at various elevated temperatures

k313

k333k353k373

Ln[D]OrLog [D](1/2.303)

Time

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47

• 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.


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• Arrhenius Plot

• -1

-2

-3Log k -4

-5

-6 1 2 3 4 5 6

1/T (°K) x 10-3

Slope = -Ea 2.303.R

°C °K 1/k40 313 3.19 x 10-3 60 333 3.00 x 10-3 80 353 2.83 x 10-3 100 373 2.68 x 10-3

k373

k353

k333

k313

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• 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.


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• The Arrhenius Equation can be re-arranged into several useful forms:

• log ( k2 ) = - Ea ( 1 - 1 )

( k1 ) 2.303 R ( T2 T1 )log of the ratio of k2 / k1

• or log k2 = Ea (T2 – T1)

k1 2.303 R.T2.T1

• Where k1 and k2 are the rate constants at temperature T1 and T2 in °K (°C+273).


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• Ea’s for some Common Pharmaceutical Degradation Reactions.

Compound Reaction Ea (kcal/mol)

Ascorbic AcidAspirinAtropineBenzocaine ChloramphenicalEpinephrineProcaineThiamine

OxidationHydrolysisHydrolysisHydrolysis

HydrolysisOxidationHydrolysisHydrolysis

2314141920231420

Range of Ea for pharmaceuticals is generally between14 - 23 kcal/mol

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• 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.


53

HO

SO2 NHCOCH3H2N Sulphacetamide

SO2NH2H2N Sulphanilamide + CH3COOH

OH-

H2N SulphacetamideSO2 N- COCH3

Ionised - Stable+H+

ionised in weak alkali

degraded in strong alkali


54

• k120 at 120°C = 9 x 10-6 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.


55

• 1. Calculate k2 ie k at 25°C

• log k25 = - Ea ( 1 - 1 )

k120 2.303 R (T2 T1 )

• log k25 = - 22 900 ( 1 - 1 )

k120 2.303 x 1.987 (298 393)

= -4.06 -ve / +ve NB

• Therefore k25 = 8.7 x 10-5 (antilog of -4.06)

k120

• Therefore k25 = 8.7 x 10-5 x 9 x 10-6 S-1

= 7.85 x 10-10 S-1


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• Now t90 = 0.105 First Order

k25

• Therefore t90 = 0.105 S-1

7.85 x 10-10

t90 = 1.34 x 108 S

t90 = approx. 4.25 years.

• This agrees with the monograph shelf life of 4.2 years at 25°C

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What percentage would degrade during the sterilisation process at 120°C for 30 mins?

Will it be significant?

k120 at 120°C = 9.1 x 10-6 S-1 first order

t = 30min = 1800 s

Ln [D]t = Ln [D]0 – k1 t

Ln [D]t - Ln [D]0 = – 9.10-6 x 30 x 60 units NB

Ln (D/D0) = - 0.01638 -ve NB

D/D0 = 0.9838

Therefore % degraded = 1.62%


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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.

• i.e.

• However, the Q10 value depends on the activation energy of the reaction.

HO

Ln[D]

Time

310°

300°


59

• 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:

EaKcal/mol

Q10 ΔT Δk

14 2 +10°C X2 Doubles

19 3 +10°C X3 Triples

24 4 +10°C X4 Quadruples

HO

Ln[D]

Time

310°

300° Ln[D]

Time

310°

300°Ln[D]

Time

310°

300°14 1924


60

Ea = 14 Kcal/mole

Ea = 19 Kcal/mole

Ea = 24 Kcal/mole

40°C313°K

30°C303°K

Q10 = k40 = 2 k30

Q10 = k40 = 3 k30

Q10 = k40 = 4 k30

3.20 3.30 3.40 1/T (°K) x 10-3

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• If temperature change is more than +10°C then:

• If ΔT = 25°C ΔT= 25 = 2.5 10 10

Therefore

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62

• If storage temperature is increased above the stated storage temperature then:

EaKcal/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 +25 2.5 15.6(32.5)

x 15.6 1/15.6

24 4 +25 2.5 32(42.5)

x 32 1/32

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63

• If storage temperature is decreased below the stated storage temperature then:

• ΔT = -25°C ΔT= -25 = -2.5 10 10

• Therefore

EaKcal/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 -25 -2.5 1/15.6(2-2.5)

x 0.064 x 15.6

24 4 -25 -2.5 1/32(2-2.5)

x 0.031 x 32

HO


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• 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


65

• 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


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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.

HO


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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.

0 18 hrs at 25°C0 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 hoursUse 72 hours as a conservative estimate (decrease temp. – use 2)


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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. 0 24 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 hoursUse 1.5 hours as a conservative estimate (increase temp. – use 4)


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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.0 12 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)


so since so since

Therefore X = 4 months Therefore X = 16 monthsUse 16 months as a conservative estimate so the product should be discarded.


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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.0 14 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)


so since so since

Therefore X = 2 days Therefore X = 8 daysUse 8 days as a conservative estimate so the product can be stored at 5°C for a further 6 days.


71

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

•


72

Transition State Theory

ΔG‡ Influences Rate

ΔG-1‡

ΔG0

ΔG‡ uncat.

ΔG0

Influences Extent

M‡ uncatalysed.

Reactants

Products

M‡ catalysed.

ΔG‡ cat.

HO


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• Types of Catalysis in Pharmaceutics1. 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+

2. Specific Base CatalysisCatalyst is the solvated hydroxyl ion from a base i.e. OH- in solution. Eg NaOH Na+ + OH-

3. General Acid CatalysisCatalyst is a proton other than an hydronium ion.Any “Bronstead Acid” or compound which can donate a proton eg. NH4 +

4. General Base CatalysisCatalyst is a proton acceptor other than OH- such as a “Bronstead Base” which acts as a proton acceptor by sharing an electron pair.


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• 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.


75

• 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 temperatureionic strength of the solutionsolvent 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

Log K

pH


76

• 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

HO


77

• V-GraphsObtained 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 catalysis2 Catalysis by 0H- - reaction proceeds under specific base catalysis3 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


78

• 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


79

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 + k2 + k3 [0H-]m

• Or k = k1 [H+]n + k2 + k3 . Kw m

[H+]

Since Kw = [H+][OH-] = 1.0 x 10-14 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


80

• Now Consider if the pH of the reacting solution is low:H+ is high soonly first term is significantthe 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


81

• Now Consider if the pH of the reacting solution is high:OH- is high soonly third term is significantthe 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.

HO


82

• 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 + k2 + 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 14


83

• 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 .

HO


84

• 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.


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• Sigmoidal GraphsThis 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.


86

• Hypothetical rate equation israte = 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.


87

• 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

100% 100% HD D-

% HD %D -

0% pKa 0%

1 pH 14


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Ionisation Profile HD H+ + D-

HO

HD H+ + D-

k† k‡ P P

If k† >> k‡

k†

Log k Log k

pKa

1 pH 14 Max HD Max D-

Min D- Min HD


89

HO

HD H+ + D-

k† k‡ P P

If k† > k‡

k†

Log k Log k

pKa


Min D- Min HD


90

HO

HD H+ + D-

k† k‡ P P

If k† = k‡

k†

Log k Log k

pKa


Min D- Min HD


91

HO

HD H+ + D-

k† k‡ P P

If k† < k‡

k‡

Log k Log k

pKa


Min D- Min HD


92

HO

HD H+ + D-

k† k‡ P P

If k† << k‡

k‡

Log k Log k

pKa


Min D- Min HD


93

HO


94

• 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

100% 100% BH+ B

% BH+ % B

0% pKa 0%

1 pH 14


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Ionisation Profile BH+ B + H+

HO

BH+ B + H+

k† k‡ P P

If k† >> k‡

k†

Log k Log k

pKa

1 pH 14 Max BH+ Max B Min B Min BH+


96

HO

BH+ B + H+

k† k‡ P P

If k† << k‡

k‡

Log k Log k

pKa

1 pH 14 Max BH+ Max B Min B Min BH+


97

HO


98

• 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- 2H+ + 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

100% 100%

pKa

% Ionisation

pKa 0% 0% 1 pH 14


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Ionisation Profile H2D H+ + D- 2H+ + D- -

HO

H2D H+ + D- 2H+ + D- -

k† k‡ k‡ †

P P P

If k† << k‡ >> k‡ †

Log k Log k

1 pH 14 Max H2D Max HD- Max D- -


100

HO


101

• 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.


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• 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 ofthe reaction.

3. Calculate points on the pH-rate profile andplot 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°C35°C

-4

Log k Log k sec-1

pH of Max Stability -6.6 1 pH 14


103

-4 -4

-5 -5

Log k Log k sec-1 sec-1

-6 -6

-7 -7 1 3.5 7 14 pH


104

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.


105

• Log k at pH 3.5 = -6.5 therefore k = 3.00 x 10-7 s-1

• Log k at pH 7.0 = -5.5 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.


106

• 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 1 3.5 7 14 pH


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• 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.

k3

H+

k1

OH-

H2O

H2O

kI

kII

Not catalysed significantly by H+ or OH- but ionisation degreases stability due to increased ionised moiety


108

• 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 > k1 and kII > kI

-2

-4

Log k Log k sec-1 -6

-8

-10 1 3 5 7 9 11 14 pH

Want a t90 = 4 months = 10 368 000 seconds

Since t90 = 0.105/k

k = 1x10-8 s-1

Log k = -8

acceptable range = 4.5 – 7.0


109

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.

Pharmaceutics II – Stability STABILITY-Vital component of all product development. All about...

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