+

Diffusion phenomena, Drug release and dissolutionAseel Samaro

+Importance of diffusion in pharmaceutical

sciences1. Drug release Dissolution of drugs from its dosage form.

2. Passage of gasses, moisture, and additives through the packaging material of the container.

3. Permeation of drug molecules in living tissue.

4. Drug absorption and drug elimination

+ Diffusion

Diffusion: is a process of mass transfer of individual molecules of a substance as a result of random molecular motion.

The driving force for diffusion is usually the concentration gradient.

+Define Diffusion

The movement of molecules from a area in which they are highly concentrated to a area in which they are less concentrated.

+Diffusion

The passage of matter through a solid barrier can occur by:

1. Simple molecular permeation or

2. By movement through pores and channels

+

+

Molecular diffusion or permeation through nonporous media depends on the solubility of the permeating molecules in the bulk membrane

The passage of a substance through solvent-filled pores of a membrane and is influenced by the relative size of the penetrating molecules and the diameter and shape of the pores

Diffusion or permeation through polymer strands with branching and intersecting channels. Depending on the size and shape of the diffusing molecules, they may pass through the tortuous pores formed by the overlapping strands of polymer. If it is too large for such channel transport, the diffusant may dissolve in the polymer matrix and pass through the film by simple diffusion.

+ Pharmacokinetics of drugs

(ADME)

Are studies of

Absorption

Distribution

Metabolism

Excretion of drugs

+ Drug Absorption and Elimination

Passage of Drugs Through Membranes

Passive diffusion

Transcellular diffusion (through the lipoidal bilayer of cells)

Paracellular diffusion (passage through aqueous channels)

Using membrane transporters

Facilitated diffusion (energy independent)

Active transport  (energy dependent)

+

+

+ How to get other molecules across membranes??There are three ways that the molecules typically move through the membrane:

1. Facilitated transport 2. Passive transport3. Active transport

• Active transport requires that the cell use energy that it has obtained from food to move the molecules (or larger particles) through the cell membrane.

• Facilitated and Passive transport does not require such an energy expenditure, and occur spontaneously.

+Membrane Transport MechanismsI. Passive Transport

Diffusion- simple movement from regions of high concentration to low concentration

Osmosis- diffusion of water across a semi-permeable membrane

Facilitated diffusion- protein transporters which assist in diffusion

+ Membrane Transport MechanismsII. Active Transport

Active transport- proteins which transport against concentration gradient.

Requires energy input

+

Elementary Drug Release

+ Osmosis

The diffusion of water across a selectively permeable membrane. Water moves from a high concentration of water (less salt or sugar dissolved in it) to a low concentration of water (more salt or sugar dissolved in it). This means that water would cross a selectively permeable membrane from a dilute solution (less dissolved in it) to a concentrated solution (more dissolved in it).

+Ultrafiltration and Dialysis

Ultrafiltration is used to separate colloidal particles and macromolecules by the use of a membrane.

Hydraulic pressure is used to force the solvent through the membrane, whereas the microporous membrane prevents the passage of large solute molecules

+ DialysisSeparation process based on unequal rates of passage of solute and solvent through microporous membrane, carried out in batch or continuous mode.

HemodialysisUsed in treating kidney malfunction to rid the blood of metabolic waste products (small molecules) while preserving the high molecular weight components of the blood

+Hemodialyzer

+ Steady-State DiffusionThermodynamic Basis

+

Fick’s laws of diffusion

Fick’s first law: The amount, M, of material flowing through a unit cross section, S, of a barrier in unit time, t, is known as the flux, J:

dtS

dMJ

dx

dCDJ Where: J is flux (g/cm2.sec)

M is the amount of material flowing (g)S is cross sectional area of flow (cm2)t is time (sec)D is the diffusion coefficient of the drug in cm2/secdC/ dx is the concentration gradient C concentration in (g/cm3)X distance in cm of movement perpendicular to the surface of the barrier

The flux, in turn, is proportional to the concentration gradient, dC/dx:

equation (1)

equation (2)

Fick’s first law

• The negative sign of equation signifies that diffusion occurs in a direction opposite to that of increasing concentration.

• That is, diffusion occurs in the direction of decreasing concentration of diffusant; thus, the flux is always a positive quantity.

• The diffusion coefficient, D it does not ordinarily remain constant.• D is affected by concentration, temperature, pressure, solvent

properties, and the chemical nature of the diffusant.• Therefore, D is referred to more correctly as a diffusion coefficient

rather than as a constant.

dx

dCDJ

dtS

dMJ

Rate of diffusion through unit area

+One often wants to examine the rate of change of diffusant concentration at a point in the system. An equation for mass transport that emphasizes the change in concentration with time at a definite location rather than the mass diffusing across a unit area of barrier in unit time is known as Fick's second law.

The concentration, C, in a particular volume element changes only as a result of net flow of diffusing molecules into or out of the region. A difference in concentration results from a difference in input and output.

Fick’s laws of diffusion

dx

dJ

dt

dC Where: J is flux (g/cm2.sec)

M is the amount of material flowing (g)S is cross sectional area of flow (cm2)t is time (sec)D is the diffusion coefficient of the drug in cm2/secdC/ dx is the concentration gradient (g/cm4)C is concentration

The concentration of diffusant in the volume element changes with time, that is, ΔC/Δt, as the flux or amount diffusing changes with distance, ΔJ/Δx, in the x direction

change in concentration of diffusant with time at any distance

dx

dCDJ

2

2

2

2

dx

CdD

dt

dC

dx

CdD

dx

dJ

Fick’s second law:

Differentiating the equation

Fick’s laws of diffusion Flux: is the rate of flow of molecules across a given surface.

Flux is in the direction of decreasing concentration.

Flux is always a positive quantity

Flux equal zero (diffusion stop) when the concentration gradient equal zero.

Diffusion coefficient also called diffusivity. It is affected by: Chemical nature of the diffusant drug. Solvent properties. Temperature Pressure Concentration

Fick’s first law

rate of diffusion through unit area

Fick’s second law

change in concentration of diffusant with time at any distance

dx

dCDJ

dtS

dMJ

2

2

dx

CdD

dt

dC

We want to calculate: • dMt/dt = release rate

• Mt = amount released after time t

+ Steady state condition With time the concentration of the diffusant molecule in the barrier

increases gradually until it reaches a steady state condition.

At the steady state at each there no change in the concentration of the diffusant with time inside the barrier.

constant

0

0

2

2

2

2

x

C

dx

dCdx

Cd

dx

CdD

dt

dC

+

Concentration will not be rigidly constant, but rather is likely to vary slightly with time, and then dC/dt is not exactly zero. The conditions are referred to as a quasistationary state, and little error is introduced by assuming steady state under these conditions.

02

2

dx

CdD

dt

dC

+ Diffusion Through MembranesSteady state diffusion through a thin film with thickness =h

h

CCDJ

dx

dCDJ

21

02

2

dx

CdD

dt

dC

Integrating the equation using the conditions that at z = 0, C= C1 and at z = h, C = C2 yields thefollowing equation:

+ Steady state diffusion through a thin film with thickness =h

R

CCJ

RD

hh

CCDJ

21

21

Where: R is diffusional resistance

dx

dCDJ

dtS

dMJ

+

Diffusional release system: Reservoir system Diffusion

+ Membrane permeability

The membrane can have a partition coefficient that affects the concentration of the diffusant inside it.

Therefore the concentration inside the membrane is a function of the concentration at the boundary and the partition coefficient of the membrane.

+ Membrane permeability

h

CCD

dtS

dMJ

21

Sink conditions cr=0

C1 : Conc. in the memb. at the donor sid C2 : Conc. in the memb. at the receptor side

h

CdDSK

dt

dM

Crh

CrCdDSK

dt

dMCr

C

Cd

CK

0

21

Rate of transport

Cumulative amount of drug released through membrane??

+

Sink Conditions : concentration of Cr is zero When? Rate of exit of drug > rate of entry

(no accumulation)

Sink Conditions

h

CdDSK

dt

dM

Crh

CrCdDSK

dt

dM

0

Membrane permeability

+ Membrane permeability

PSCddt

dMRh

DKP

h

CdDSK

dt

dM

1 Where P is the permeability of the membrane in cm/sec.

RD

h

Where: R is diffusional resistance

P = permeability coefficient (cm/s)

+ Membrane permeability

tPSCdM

Cd

PSCddt

dM

constant

M: is the amount of diffusant that passes through the membrane after time t.

Cumulative amount of drug released through membrane

+

Zero-order process

- Amount of drug transported is constant over time- Only if Cd does not change- Diffusion of drug from transdermal patch

tPSCdM

Diffusion

Diffusion

+ Example : To study the oral absorption of paclitaxel(PCT) from an oil-water emulsion

formulation, an inverted closed-loop intestinal model was used.

- surface area for diffusion = 28.4 cm2

- concentration of PCT in intestine = 1.50 mg/ml.

- the permeability coefficient was 4.25 x 10-6 cm/s

Calculate the amount of PCT that will permeate the intestine in 6 h of study (zero-order transport under sink conditions)

Using the equation M= PSCdt ,

M = (4.25 x 10-6)(28.4)(1.50)(21,600) = 3.91 mg

+First-order transport

If the donor conc. changes with time,

(Cd)t : donor conc. at any time(Cd)0 : initial donor conc. Vd : volume of the donor compartment (mL)

Diffusion

tVd

PSCdCdt 303.2

)0(loglog

d

dd V

MC

PSCddt

dM

tVd

PSCdCdt )0(lnln

+

Con

c. o

f di

ssol

ved

drug

Time

First order dissolution under non-sink condition

Zero order dissolution under sink condition

Dissolution rate

Fig. Butyl paraben diffusing through guinea pig skin from aqueous solution. / H. Komatsu and M. Suzuki, J. Pharm. Sci., 68 596 (1979)

+ Burst effect

In many of the controlled release formulations, immediately upon placement in the release medium, an initial large bolus of drug is released before the release rate reaches a stable profile. This phenomenon is typically referred to as ‘burst release.’

Initial release of drug into receptor side is at a higher rate than the steady-state release rate

+

Lag time, Burst effects

+Lag time, Burst effects

Lag time : time of molecules saturating the membrane

tL = h2 / 6D h : membrane thickness (cm)

D : diffusion coefficient (cm2/s)

P

htL 6

)( Ld

d

tth

DSKCM

th

DSKCM

h

CdDSK

dt

dM

+Lag time, Burst effect

Burst effect : time of initial rapid release of drug

tB = h2 / 3D h : membrane thickness (cm) D : diffusion coefficient (cm2/s)

P

htB 3

)( B

d

d

tth

DSKCM

th

DSKCM

h

CdDSK

dt

dM

+ Example

The lag time of methadone, a drug used in the treatment of heroin addiction, at 25°C (77°F) through a silicone membrane transdermal patch was calculated to be 4.65 min. The surface area and thickness of the membrane were 12.53 cm2 and 100 µm, respectively.

a. Calculate the permeability coefficient of the drug at 25°C (77°F) (K = 10.5).

b. Calculate the total amount in milligrams of methadone released from the patch in 12 h if the concentration inside the patch was 6.25 mg/mL.

+

Solution

a. To use the equation P = DK/h , the diffusion coefficient D should be determined.

Therefore, using the lag-time equation tL = h2 / 6D

D= h2 / 6 tL

= (1.00 x 10-2)2 / (6)(279)

= 5.97 x 10-8 cm2/s

Therefore,

P = DK / h

= [(5.97 x 10-8)(10.5) / (1.00 x 10-2)

= 6.27 x 10-5 cm/s

Diffusion o The surface area 12.53 cm2

o thickness of the membrane 100 umo lag time, 4.65 min and K = 10.5

+

Solution

Using the equation M = PSCd(t - tL),

M = [(6.27 x 10-5)(12.53)(6.25)][(43,200) – (279)]

= 210.8 mg

Diffusion

b. Calculate the total amount in milligrams of methadone released from the patch in 12 h if the concentration inside the patch was 6.25 mg/mL.

+ Fick’s first law dx

dCDJ

dtS

dMJ

Fick’s second law 2

2

dx

CdD

dt

dC

Diffusion Through Membraneswith thickness =h h

CCDJ

21

Sink Conditions h

CdDSK

dt

dMh

CrCdDSK

dt

dM

Rate of transport

Rh

DKP

1

Membrane permeability

+PSCd

dt

dM

Zero-order process

tPSCdM

tconsCd tan

First-order transport

tVd

PSCdCdt 303.2

)0(loglog

tVd

PSCdCdt )0(lnln

+Lag time Burst effects

tL = h2 / 6D

M = PSCd(t - tL)

tB = h2 / 3D

M = PSCd(t + tB)

+ Multilayer Diffusion Diffusion across biologic barriers

The passage of gaseous or liquid solutes through the walls of containers and plastic packaging materials

The passage of a topically applied drug from its vehicle through the lipoidal and lower hydrous layers of the skin.

+ Multilayer Diffusion

+ Multilayer Diffusion

i

iii h

KDP

ii

i

ii KD

h

PR

1

The total resistance, R

nRRRR .......21nPPPP

1.....

111

21

The total permeability for the two layers

112221

2211

KDhKDh

KDKDP

+ Procedures and Apparatus For Assessing Drug Diffusion