Estabilidad

Estabilidad Estabilidad
Tecnologia Farmacéutica. Tecnologia Farmacéutica.

Estabilidad de Medicamentos Contenido en sustancia activa

Caracteristicas esenciales de calidad: Periodo de validez:
− Contenido en sustancia activa − reducción del 10% en sustancia activa
− Forma farmacéutica y caracteres − Toxicidad no aumenta debido a productos de
organolépticos degradación
− Flora microbiana Fecha de caducidad: mes y año
− Toxicidad Periodo de validez máximo 5 años (RD
− Biodisponibilidad 736/1987 de 17 de marzo)

Estables, seguros, eficaces

Depto. Farmacia y Depto. Farmacia y
1 2
Tecnología Farmacéutica Tecnología Farmacéutica
© Marival Bermejo © Marival Bermejo


Estudios de estabilidad Alteraciones de los medicamentos

Sobre la sustancia activa Inestabilidad Física
Fase de preformulación: Inestabilidad Química
− Sólido y disolución Inestabilidad Microbiológica
− Compatibilidad con excipientes
− Estabilidad frente a operaciones básicas
Especialidad terminada
− Ensayos a largo plazo
− Ensayos de envejecimiento acelerado.

3 4

1


Inestabilidad Física: alteración de las
Inestabilidad química
caracteristicas galénicas

Consecuencias: Cinética química
− Aspecto
Orden y molecularidad de la reacción
− Regularidad de la dosificación
− Biodisponibilidad (caducidad biofarmacéutica)
Vias degradación física:
− Polimorfismo
− Crecimiento cristalino
− Sedimentación
− Floculación
− Coalescencia

5 6


Stability Calculations Stability Calculations

P
D W
a typical chemical reaction

− Reaction rate ∝ [D]*[W]
P
D W
D and W, quot;collidingquot; to form one or more product molecules,
d[D]
− α[D][W]
the rate of reaction: proportional to the number of collisions a

dt
the number of collisions is proportional to the product of the concentrations of the
two species:
− we let k2 be the rate constant
Reaction rate ∝ [D]*[W]
−
d[D]
− = k 2 [D][W]
Reaction rate corresponds to the rate of loss of D and is denoted (d[D]/dt)
dt

7 8

2


Stability Calculations Stability Calculations
Order of reaction:
Order of reaction: sum of the exponents of
the concentration terms in the rate equation d[D]
− = k 2 [D][W]
dt
− If the water concentration W is held constant (by having a
d[D]
− = k 2 [D][W] large excess, as in most solutions) then :
dt
d[D]
− − k1[D] where k1= k2 [W]
dt
Second order reaction
and the reaction is apparent first order or pseudo first order

9 10


Stability Calculations
Order of reaction:
− If, now, in addition, we also fix the drug Orden Ecuación Unidades k
concentration (e.g., by making a suspension), the velocidad
equation becomes: C*t-1
0 -dC/dt=k
d[D]
− = k0
t-1
dt 1 -dC/dt=-k*C

-dC/dt=k*C2 C-1*t-1
2
where k0 = k1[D] = k2[D][W]
-dC/dt=k*C*B
-dC/dt=k*Cn Cn-1*t-1
n
and the reaction is apparent zero order

11 12

3


First-Order Calculations

[D] = [D]0 e − k1t
D P
Orden Ecuación Ecuación
velocidad integrada
First order First order
0 -dC/dt=k C = C0 − k ⋅ t 1000.0

L n C o n c e n tra tio n
500.0

C o n c e n tra tio n
400.0
1 -dC/dt=-k*C 100.0
− k ⋅t 300.0
C = C0 ⋅ e
200.0
10.0
100.0
-dC/dt=k*C2
2 1 1
= + k ⋅t 1.0 0.0
C C0 0 100 200 300 400 500 0 100 200 300 400 500

Time (min)
Time (min)

T1/2

13 14


First-Order Calculations First-Order Calculations
The half-life, t½: The shelf-life, t90:
usually taken to be the time for [D] to reach 0.90[D], that is,
is the time for [D] to become [D]/2,
10% decomposition,

[D]0
= ln[D]0 − k1t 12
ln
0.105
2
t 90 =
0.693
ln 2 k1
t 12 =
= t 12
k1 k1

15 16

4


First-Order Calculations Zero-Order Calculations
rate equation with no concentration dependence.
Example:
aspirin (acetylsalicylic acid) d[D]
− = k0
at pH 2.5 the (pseudo-first-order) rate constant is 5 dt
x 10-7s-1 at 25ºC.
Integrating from t = 0 to t = t with [D] = [D]0 at t = 0,

half-life [ D] t

∫ d[D] = − ∫ k 0dt
0.693
t 12 = = 1.39 x 106 s
5 x 10-7 [ D]0 0
shelf life
0.105
t 90 = = 2.1 x 105 s [D] = [D]0 - k0*t
5 x 10-7

17 18


Zero-Order Calculations Zero-Order Calculations
[D] = [D]0 - k0*t
D] = [D]0 - k0*t Zero order
Half life 1000
900
800
0.5[D]0
t 12 = 700

Concentration
600
k0 500

Shelf-life 400
300
200
100

0.1[D]0 0

t 90 = 0 20 40 60 80
Time mins
k0
0.5[D]0
t 12 =
k0
19 20

5


Zero-Order Calculations TEMPERATURE EFFECTS
Example: aspirin (acetylsalicylic acid)
at pH 2.5 the (pseudo-first-order) rate constant is
Activation Energy Calculations
5 x 10-7s-1 at 25ºC.
Reaction rates are proportional to the number of collisions per unit time.
Aspirin solubility 0.33 g/100 mL.
for an aspirin suspension: zero-order rate the number of collisions increases as the temperature increases.

k0 = 5 x 10-7s-1 x 0.33 g/100 mL = The reaction rate constant is observed to have an exponential dependence on
temperature described by the Arrhenius equation
k0 = 1.65 x 10-7 g/(100mL * s)
k = A exp(-Ea/RT)
Dose of aspirin: 650 mg/teaspoon(5 mL) = 13 g/100 mL.
− k :reaction rate constant
− A :constant
shelf life − Ea: activation energy of the chemical reaction
− T : the absolute temperature (t ºC + 273.16ºC).
(0.1)(13)
t 90 = = 7.9 x 106 s=91 days log k = log A - Ea/2.303RT
-7
1.65 x 10

21 22


TEMPERATURE EFFECTS
TEMPERATURE EFFECTS Typical Arrhenius plot of log k against 1/T,
60

4.5
50 4
3.5
Energy Kcal/mol

Eact= 12 Kcal/mol 3

Log K
40
2.5
2
30 1.5
A+B 1
reactants 0.5
20
0
3 3.2 3.4 3.6 3.8
B+C
10
(1/T(Kº))*1000
products

Slope= 3.51*103
0
Ea= 3.51*103*(2.303)*(1.987)=16.1 Kcal/mol
Extent of reaction

23 24

6


TEMPERATURE EFFECTS
TEMPERATURE EFFECTS
Example:
Compound Reaction Ea (kcal/mol)
Sulfacetamide.
Ascorbic Acid Oxidation 23
first-order rate constant in the pH-independent region (5-11)
Aspirin Hydrolysis 14
K= 9 x 10-6 s-1 at 120ºC.
Atropine Hydrolysis 14 Ea= 22.9 kcal/mol at pH 7.4.
Calculate the shelf life at 25ºC.
Benzocaine Hydrolysis 19
Chloramphenicol Hydrolysis 20
−22900 ⎛ 1 1⎞
− Ea ⎛ 1 1 ⎞ k 25
k2
Epinephrine Oxidation 23
= −
= ⎜−⎟ ⎜ ⎟
log
log
k120 (2.3)(1.987) ⎝ 298 393 ⎠
k1 2.303R ⎝ T2 T1 ⎠
Procaine Hydrolysis 14
Thiamine Hydrolysis 20
k 25
= 8.7 x 10-5 k25 = (8.7 x 10-5) (9 x 10-6 s-1)
= 7.85 x 10-10 s-1
k120
Activation Energies for Some Drug Decomposition Reactions
0.105
t 90 = = 1.34 x 108 s 4.25 years
7.85 x 10-10 s-1
25 26


TEMPERATURE EFFECTS TEMPERATURE EFFECTS
Example Answer: (a) from the Arrhenius plot of the data we obtain
The following rate constants were determined for 5-fluorouracil The slope
decomposition at pH 9.90.
B= -12.31=Ea/R
Ea = 24.5 kcal/mol
106 k (s -1 )
t (ºC)
80 0.96
10
70 0.32
60 0.118 1

Log k
0.1
Determine the activation energy at this pH;
(a)

Extrapolate the graph to room temperature (25ºC) and
(b)
determine the rate constant and shelf life at this temperature. 0.01
2.800 2.850 2.900 2.950 3.000 3.050
1/T (Kº)

27 28

7


TEMPERATURE EFFECTS Q10-Value Calculations:
(b) From extrapolation of the graph to 25ºC (1/T = 0.00335),
Consider the ratio of rate constants kT2/kTl at two temperatures
we obtain log k25ºC = -8.86. k25ºC = 1.38 x 10-9 s-1
T1 and T2. Consider T1=T2+10º
t90 = 7.6 x 107 s = 2.4 years

The quantity Q10 is defined as
10

k (T +10)
Q10 =
1 kT
Log k The Q10 factor can be calculated from the next equation:
0.1

⎡E 1 ⎞⎤
⎛1
k (T +10)
Q10 = = exp ⎢ − a −⎟
0.01
⎜
T + 10 T ⎠ ⎥
2.800 2.850 2.900 2.950 3.000 3.050
⎝
⎣R ⎦
kT
1/T (Kº)

Note that Ea is always positive.

29 30


Q10-Value Calculations: Q10-Value Calculations:
For an arbitrary change in temperature, ΔT = T2 - T1,
The activation energies for drug decompositions usually fall in the
range 12 to 24 kcal/mol.
Next table gives values of Ea corresponding to three rounded
values of Q10.
These values of Q10 = 2, 3, or 4 to represent low, average, and
high estimates of Q10 when Ea is unknown k (T +ΔT)
= Q10( ΔT /10)
Q ΔT =
kT
Q10 (30 to 20ºC) Ea (kcal/mol)
2.0 12.2
3.0 19.4
4.0 24.5

31 32

8


For an arbitrary change in temperature, ΔT = T2 - T1,
⎡E ΔT (T + 10) ⎤
10
Q ΔT = exp ⎢ a
R (T + 10)(T) 10 (T + ΔT) ⎥
⎣ ⎦
⎡E ⎛ ⎞⎤
ΔT
Q ΔT = exp ⎢ a ⎜ ⎟⎥ [ ΔT /10][(T +10) /(T +ΔT )]
⎣ R ⎝ (T + ΔT)(T) ⎠ ⎦ ⎡E ⎤
10
= exp ⎢ a ⎥
⎣ R (T + 10)(T)1 ⎦
Multiplying the exponential term by

Q ΔT = Q10[ ΔT /10][(T +10) /(T +ΔT )]
10 T + 10
i
10 T + 10 This equation is exact. However, since T ≈ 300K, (T + 10)/(T + ∆T) ≈ 1,

gives ⎡E 10 (T + 10) ⎤
ΔT k (T +ΔT)
Q ΔT = exp ⎢ a
Q10 ΔT /10
⎥
Q ΔT =
⎣ R (R + ΔT)(T) 10 (T + 10) ⎦
kT
33 34


Example Calculate the factors by which rate constants may (b) a 25 to 0ºC temperature change.
Answer:
change for:
(a) a 25 to 50ºC temperature change
(b) a 25 to 0ºC temperature change. ΔT = -25,
Answer:
k (0−25)
= Q10 ( −25/10)
Q−25 =
(a) ΔT = +25,
k0
k (25+ 25)
Q+25 = = Q10 (25/10)

k 25 = 1/5.7, 1/15.6, 1/32 for Q10 = 2, 3, 4, respectively

Thus the rate decreases to between 1/6 and 1/32 of the initial
= 5.7, 15.6, 32 for Q10 = 2, 3, 4, respectively rate.

Thus the rate increases between 6-fold and 32-fold, with a
probable average increase of about 16-fold.

35 36

9


SHELF-LIFE ESTIMATION METHODS
Q10-Value Calculations: The expiration date is given for room temperature. What is
• Q = 4 provides the largest estimate for the increase in rate the expected extension of the shelf life in a
with increasing temperature, refrigerator?
• Q = 2 provides the smallest reasonable estimate for the
decrease with decreasing temperature.

Example: Using Q10 = 2, 3, 4, calculate the half-life change from
70ºC to 25ºC for methyl paraben at pH 4.
We have ΔT = 25 - 70 = -45,
Q-45= 1/23, (Q10=2)
Q-45= 1/140, (Q10=3)
Q-45= 1/512, (Q10=4)

Since k70ºC = 1.6 x 10-6 s-1 (from the monograph) then
t½ (70ºC) = 4.3 x 107 s = 501 days

t½ (25ºC) = 31 yr (Q =2)
= 199 yr (Q = 3)
The expiration date is given for refrigeration conditions.
= 709 yr (Q = 4)
How long may the product be left at room temperature?
.
37 38


The expiration date is for room-temperature conditions, and it
is desired to heat the product in sterilization. What percent
decomposition can be expected at the higher temperature?
The expiration date is for room-temperature conditions, and it
is desired to heat the product in sterilization. What percent
decomposition can be expected at the higher temperature?

0.1[D 0 ]
Zero order t 90 =
k0
a
t 90 (T) =
k (T1 +ΔT)
0.105
First order t 90 =
k1
Problems of this type require estimates of the effect of
temperature on the shelf life. Assuming we do not have
exact Ea values available, we use the Q10 values of 2, 3,
and 4 to make such estimates.

39 40

10


Example A.
a The expiration period for a reconstituted product is 18 h at room
t 90 (T2 ) = temperature.
k T1 iQ10 ( ΔT /10)
Estimate the expiration period when the product is stored in the refrigerator.
or, since t90(T1) = a/kT1:
ΔT = -20ºC,
18
t 90 (T1 ) t 90 (5º ) =
t 90 (T2 ) =
Q10 ( −20 /10)
t (T ) Q10 ( ΔT /10)
t 90 (T2 ) = 90( ΔT 1
Q10 /10) t90 (5ºC) = 18 * 22 = 72 h (Q = 2)
= 18 *32 = 162 h (Q = 3)
the estimate of t90(T2) is independent of the order. = 18 * 42 = 288 h (Q = 4)
a positive ΔT (T2 > T1) reduces the shelf life,
a negative ΔT(T2 < T1) increases it. conservative estimate would be 72 h and a likely estimate 162 h.

41 42


Example B. If the product has been stored for a known length of time
A newly reconstituted product is labeled to be stable for 24 h in a at another temperature:
refrigerator. What is the estimated shelf life at room temperature?
compute the time interval at the specified
1.
t90 (5ºC) = 24 h
temperature that would give the equivalent
24
t 90 (25º ) =
t 90 (T1 )
t 90 (T2 ) = decomposition to that which occurred at the actual
Q10 (20 /10)
Q10 ( ΔT /10) storage temperature and
subtract this value from or add it to the label date for
2.
t90 (25ºC) = 24/4 = 6 h (Q = 2)
a new expiration date.
= 24/9 = 2.7 h (Q = 3)
= 24/16 = 1.5 h (Q = 4)
Thus a likely value is about 3 h, with a conservative estimate of
1.5 h and a possible value of 6 h.

43 44

11


Example Example.
The expiration date for a product is one year from the current date when
The ampicillin monograph in the Physicians' Desk Reference (1) states
stored in a refrigerator. The product has been stored-for one month at
that the reconstituted suspension is stable for 14 days in a
room temperature. If the product is now returned to the refrigerator,
refrigerator. If the product is left at room temperature for 12 h, what
what is the new expiration date?
is the reduction in the expiration period?
Answer
One month at room temperature would be equivalent to
t 90 (T1 )
→ t 90 (T2 ) ⋅ Q10 ( ΔT /10)
t 90 (T2 ) =
t 90 (T1 )
→ t 90 (T2 ) ⋅ Q10 ( ΔT /10) Q10 ( ΔT /10)
t 90 (T2 ) =
Q10 ( ΔT /10)
0.5 * 22 = 2 days (Q = 2)
1 *22 = 4 months (Q = 2) 0.5 * 32 = 4.5 days (Q = 3)
1 *32 = 9 months (Q = 3)
0.5 * 42 = 8 days (Q = 4)
1 *42 = 16 months (Q = 4)
that is, 4, 9, or 16 months at 5ºC.
The expected reduction is thus 4.5 days.
The most likely estimate is 9 months, hence if the product is returned to the
refrigerator, only 3 months would be left.

45 46


Influencia del pH Influencia del pH

Example1

pH-Rate profile for hydrolysis of citric acid anhydride at 25°C
Example:pH-Rate profile for the dehydration of streptovitacin A at 70°C
47 48

12


D. pH EFFECTS (2). Sigmoid Curves D. pH EFFECTS (3). Bell-Shaped Curves
Example2: Hydrolysis of aspirin.
Some pH-rate profiles show
maxima, often with a quot;bell
where k3 > kl and kquot; > k'.
shapedquot; peak.
The rate equation is

rate = k1[RCOOH][H+] +
k'[RCOOH] + kquot;[RCOO-] +
k3[RCOO-][OH-]
The pKa of aspirin is 3.6, so the
plateau region of the curve (pH 4
to 8) is accounted for primarily by
Hydrolysis of penicillin G catalyzed
the kquot; term, that is, hydrolysis of
by 3,6-bis (dimethylaminomethyl)-
the anion. This reaction is
catechol.
responsible for the instability of
aspirin in solution dosage forms
Catalysis is produced by the monoanion of the catechol derivative,
pH-Rate profile for aspirin hydrolysis at 25°C
49 50


Vías de degraddación química Inhibición de la oxidación

Hidrólisis Proteger de la luz
Oxidación Evitar presencia oxígeno
Racemización Uso agentes antioxidantes
Descarboxilación
Polimerización
Descomposición enzimática

51 52

13


Antioxidantes sistemas hidrófilos Antioxidantes sistemas lipófilos

Combinaciones inorgánicas de azufre: Naturales: Tocoferoles
sulfitos (olor y sabor desagradables) Sintéticos y semisintéticos:
−
Combinaciones orgánicas de azufre: (olor Ácido gálico y ésteres del ácido gálico
−
y sabor desagradables) Ésteres ácido ascórbico
− Butilhidroxianisol(BHA)
Ácido ascórbico: tópicas, parenterales,
−
orales Butilhidroxitolueno(BHT)

53 54


Sinérgicos Sobredosificación

Reforzadores de acción: ácido fosfórico, Antibióticos, vitaminas
citríco, tartárico Exceso de contenido activo que es
Inactivadores de catalizadores: EDTA necesario tener en cuenta para garantizar
la estabilidad, es decir, la compensación
de la pérdida de principio activo prevista
durante el periodo de conservación.

55 56

14


Inestabilidad biológica Clasificación conservantes

Conservadores: requisitos Fenoles
Tolerancia fisiológica Alcoholes alifáticos y aromáticos
Compatibilidad Compuestos orgánicos de mercurio
Estabilidad química Compuestos de amonio cuaternario
Olor y sabor Ácidos carboxílicos
Espectro activo Otros.

57 58

15

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