Unit 1: Solubilization of Drugs in Aqueous Media

Introduction:

1. Solution: A solution is a mixture of two or more substances, e.g., sugar dissolved in water.

2. In sugar solution, sugar is called solute and water is solvent.

3. Saturated solution: A solution which is saturated with solute is called saturated solution.

Example: 100 ml of water can dissolve 36 g of salt at 20 OC.

Example: 100 ml of water can dissolve 200 g of sugar at 20 OC.

Example: 100 ml of water can dissolve 1.2 g of paracetamol at 20 OC.

4. Solubility: The concentration of solute in a saturated solution is called solubility.

Example: Solubility of paracetamol is 1.2 % in water at 20 OC.

5. Unsaturated solution: When the concentration of solute is less than saturation solubility, the

solution is an unsaturated solution. Example: 10 g salt dissolved in 100 ml water.

6. Super saturated solution: When the concentration of solute is more than saturation solubility, the

solution is a super saturated solution. Example: 38 g salt dissolved in 100 ml water at higher

temperature and is slowly cooled to 20 OC.

Solubilization:

1. Increasing the solubility of drugs in aqueous media (water) is called solubilization.

2. There are various methods of solubilization: Solid state manipulation, pH adjustment; co solvent

addition; micellar solubilization and complexation.

3. The method selected depends on efficiency, stability of drug in the system and bio compatibility

with delivery route.

Solubility:

1. The concentration of solute in a saturated solution is called solubility. Example: Solubility of

paracetamol is 1.2 % in water at 20 OC or 1 g in 70 ml water..

2. Crystal structure (melting point) and molecular structure (activity coefficient / octanol water

partition coefficient) of the drug affect its solubility in water.

3. The aqueous solubility of an organic solute is given by the below equation.

log Xw = log Xi – log γw ---------- 1

Xw = solubility of organic solute / drug in water

Xi = ideal solubility of organic solute / drug in water

γw = activity coefficient of organic solute / drug in water

Ideal Solubility:

1. The ideal solubility depends on the crystalline structure of solute.

2. A solute molecule must first dissociate (break away) from crystal lattice structure to go into

solution. This dissociation from the crystalline structure requires energy. If more energy is

required to free a solute molecule from its crystal (i.e., the higher the melting point), the lower

will be the solubility.

3. The ideal solubility can be calculated from the below equation.

Xi = - 0.01 ( M.P – 25) ---------- 2

4. Where Xi is the ideal solubility (mole fraction), M.P is melting point of solute in centigrade.

5. If the melting point is less than 250 C, the solute is a liquid at room temperature, and there is no

crystalline effect on solubility. The above equation becomes zero.

6. For crystalline compounds, the ideal solubility is strictly a function of the pure crystal, and as a

result, is solvent independent.

7. The below table gives the effect of melting point on solubility. It can be seen that for every

100 degree difference in melting point, there is a corresponding change in solubility of 10 times.

S.NO Hypothetical Melting Point in 0C Decrease in Solubility

1 0 No effect, solute is liquid

2 25 No effect, solute is liquid

3 125 10 X

4 225 100 X

5 325 1000 X

Activity Coefficient:

1. The activity coefficient of a solute describes the effect of molecular structure on solubility.

2. If the solute mixes with water readily and forms an ideal solution, the activity coefficient is taken

as unity and the last term in equation 1 becomes zero. In such cases, the crystalline structure is

the only property influencing the aqueous solubility of the drug.

3. The aqueous solubility of an organic solute is given by the below equation.

log Xw = log Xi – log γw ---------- 1

Xw = solubility of organic solute /drug in water

Xi = ideal solubility of organic solute /drug in water

γw = activity coefficient of organic solute /drug in water

4. Most drugs are non polar and do not form ideal solutions in water. Hence activity coefficient of a

drug can be used to study the influence of molecular structure on aqueous solubility.

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5. The activity coefficient can be found from octanol water partition coefficient, using the below

equation.

Ko/w = octanol – water partition coefficient of drug.

If the drug is lipophillic, it will have more partition coefficient and lesser aqueous solubility.

6. As Ko/w increases by a factor of 10, the decrease in solubility is also a factor of 10 as shown in the

below table.

log γw = - log Ko/w + 0.80 ---------- 3

S.NO Ko/wDecrease in Solubility

1 1 Reference

2 10 10

3 100 100

4 1000 1000

Estimating Solubility:

1. The concentration of solute in a saturated solution is called solubility.

2. The solubility of a drug in water depends on its crystal structure and molecular structure of the

drug molecule.

8. A solute molecule must first dissociate (break away) from crystal lattice structure to go into

solution. This dissociation from the crystalline structure requires energy. If more energy is

required to free a solute molecule from its crystal (i.e., the higher the melting point), the lower

will be the solubility.

3. Octanol – water partition coefficient will give an idea about the hydrophilic / lipophillic nature of

the drug molecule. Higher the partition coefficient, lesser the aqueous solubility.

4. From the octanol - water partition coefficient and the melting point of the drug molecule, we can

estimate the solubility of a drug using the below equation.

log Sest = -0.01 (MP – 25) – log K o/w + 0.80 --------- 4

S est = Estimated solubility of the drug

MP = melting point of drug in centigrade

K o/w = octanol – water partition coefficient of drug

5. For drugs that are liquid at room temperature, melting point term is zero and equation 4 becomes

log Sest = – log K o/w + 0.80 --------- 5

6. Equations 4 and 5 can be used to understand why a drug is poorly soluble.

7. If a compound has a low melting point and a high Ko/w, then molecular structure is limiting

solubility and the aqueous media must be modified for solubilization.

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8. If the melting point is very high, and the Ko/w is low, then modification of the aqueous media

may not significantly increase solubility and manipulation of the solid phase may be necessary.

9. If a compound has a high melting point and a high log Ko/w, it can be a challenge to the

formulator for drug solubilization.

Estimating Dissolution Rate:

1. The speed with which a drug dissolves in the solvent is called dissolution rate. It is given by the

below equation.

2. Dissolution rate or speed = Amount dissolved / time taken = dM / dt

3. Example: Time taken for 50 mg of paracetamol to dissolve in water is 10 minutes.

Speed of dissolution = 50 mg / 10 min = 5 mg / min.

This indicates that paracetamol is dissolving at a speed of 5 mg in each minute.

4. Drugs having greater solubility will have greater dissolution speed.

5. Dissolution rate of a drug can be estimated using the below theoretical equation.

6. dM / dt = K A (S - C) --------- 6

dM = amount of drug dissolved, dt = time taken for dissolution, S = solubility of drug in solvent

C = concentration of solute in solution

A = surface area of the drug exposed to dissolution media

K = a constant that includes diffusion coefficient of solute and other hydrodynamic properties

of system.

7. From the above equation, it is clear that,

a. Greater the solubility of the drug, greater will be the dissolution rate.

b. Larger the surface area (small particles have large surface area), greater the dissolution rate.

8. Oral bioavailability problems are due to poor solubility and dissolution rate of drugs in G.I fluids.

Hence bioavailability can be improved by selection of more soluble and fast dissolving forms of

drugs.

9. Example: Polymorph B of chloramphenicol palmitate has two times greater solubility than

polymorph A, and gave greater in vivo blood levels.

Hildebrand Solubility Approach / Solubility Parameter:

1. The Hilderband solubility parameter is a measure of cohesive forces between like molecules. The

greater the value, the greater the cohesive attractive forces between like molecules.

2. In a solution process

a. Attractive forces between solute molecules are broken. Energy is required for this step. Energy

consumed in this step will depend on the attractive forces between solute molecules.

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b. Attractive forces between solvent molecules are broken. Energy is required for this step. Energy

consumed in this step will depend on the attractive forces between solvent molecules.

c. Now solute molecule is given accommodation between solvent molecules, solute molecules

develop attractive forces with solvent molecules. Energy is released in this step. Energy released

in this step will depend on the attractive forces developed between solute and solvent molecules.

d. If the attractive forces in solute, solvent and solute – solvent molecules are of similar magnitude,

then dissolution will take place readily.

3. Solubility parameter is a measure of these attractive forces in solute and solvent.

4. Greater the attractive forces, greater the solubility parameter.

5. If solubility parameters of solute and solvent are nearly same, the solute will have high solubility

in that solvent.

6. Solubility parameter can be calculated from heat of vaporization, internal pressures, surface

tensions and other properties of the substance.

7. Solubility parameter of phenobarbitone is 12.6 Hilderbands (H).

8. Solubility parameter of water = 23.4 H, glycerol = 17.7 H, alcohol = 13, chloroform = 9.3 H

9. Phenobarbitone will have maximum solubility in alcohol because its solubility parameter is

matching with alcohol.

10. If necessary solvents can be mixed to get the desired solubility parameter. Alligation method can

be used to calculate the rato in which the solvents should be mixed. For example, if water and

alcohol are mixed in the ratio 1:1, the solubility parameter of the blend will be

(23.4 + 13) / 2 = 18.2

11. Hence using solubility parameters, a solvent blend can be selected for a particular drug.

Apparent Solubility Enhancement From Different Solid Phases / Solid State Manipulation:

Polymorphism:

1. When a substance exists in more than one crystalline form, the phenomenon is called

polymorphism.

2. Polymorphs of a drug have different physical properties. They vary in melting point, solubility,

dissolution rate, density, hardness, compression characteristics etc. They do not vary with respect

to chemical and pharmacological properties.

3. Polymorphs can be prepared by crystallizing the drug from different solvents.

4. Out of all the polymorphs, one of the polymorph has high physical stability, high melting point

and low aqueous solubility. The remaining polymorphs have less physical stability, low melting

points and more aqueous solubility. They are called meta stable polymorphs.

5. So, meta stable forms are used in making tablets, so that the drug will dissolve quickly in G.I

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fluids and bio availability will be more.

6. Example: Riboflavine is available as form I, II and III. Form III is 20 times more water soluble

than form I.

7. Example: Mebendazole exists in three diferrent crystalline forms A, B and C. Form B has 7 times

greater solubility than form A.

Amorphous Form and Crystalline Forms:

1. A drug may exist as crystalline form or amorphous form. If a drug does not have any crystalline

form, it is called amorphous form.

2. A crystalline drug can be converted into amorphous form by melting the crystalline drug and

cooling it rapidly. The drug molecule will not be given time for forming crystals. The molecules

will be arranged randomly in the solid.

3. The amorphous form has low melting point and more aqueous solubility and dissolution rate.

4. Example: Amorphous theophylline has 58 times greater solubility than crystalline form.

5. Example: Amorphous morphine has 268 times greater solubility than crystalline form.

Anhydrates and Hydrates:

1. If a drug has water as a part of its crystalline structure, it is called hydrate.

2. The hydrate and anhydrate form of drugs have different solubilities. The anhydrates have greater

solubility than the hydrates.

3. Example: Ampicillin is available as anhydrous ampicillin, ampicillin mono hydrate and

ampicillin tri hydrate. Anhydrous ampicillin has two times greater solubility than ampicillin

hydrate.

4. If an organic solvent is present inside the crystal, it is called organic solvate. Generally, the

organic solvates have greater solubility than the drug. Example: Pentanol solvate of

glibenclamide has 31 times greater solubility than glibenclamide.

pH control :

1. Changing the pH of the solvent system can be done for increasing solubility of organic ionizable

drugs.

2. For weak acidic and weak basic drugs, the following equations can be used to predict the effect of

pH on the solubility of the drug.

3. For weak acidic drugs, ST = Sw (1 + 10 (pH – pKa) ) ---------- 7

4. For weak basic drugs, ST = Sw (1 + 10 ( pKa - pH) ) ---------- 8

ST = Total Solubility of drug , Sw = Total Solubility of drug in water

PKa = negative logarithm of dissociation constant

5. Example: A weakly acidic drug with a pKa of 5 has a solubility of 2 mg in one liter of water. Find

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out its solubility in solvents having a pH of 1 and 8.

6. For weak acidic drugs, ST = Sw (1 + 10 (pH – pKa) ) ---------- 7

7. Solubility in a solvent at pH = 1

ST = 2 (1 + 10 (1 – 5) ) = 2 (1 + 10 – 4) = 2 (1+0.0001) = 2 (1.0001) = 2.0002 mg / liter

8. Solubility in a solvent at pH = 8

ST = 2 (1 + 10 (8 – 5) ) = 2 (1 + 10 3) = 2 (1+1000) = 2 (1001) = 2002 mg / liter

9. From the above example, it is clear that the solubility of the drug has increased by 1000 times

with change in pH to alkaline side. In acidic pH, the solubility was almost same.

10. The reason for this is, weakly acidic drugs will be ionized in alkaline media and will have more

solubility. In acidic media, they remain unionized and have low solubility in water.

11. Similarly, weak basic drugs will have more solubility in acidic media and less solubility in basic

media.

12. The effect of pH on solubility of weak acidic drugs and weak basic drugs is shown in below

graphs.

Effect of pH on Solubility of a Weak Acidic Drug

0

1

2

3

4

5

6

7

8

0 2 4 6 8 10 12 14 16

pH

log

(so

lub

ilit

y)

7

Effect of pH on Solubility of a Weakly Basic Drug

0

1

2

3

4

5

6

7

0 2 4 6 8 10 12 14 16

pH

log

(so

lub

ilit

y)

Buffers:

1. Buffers are used to maintain the pH of a system.

2. Solubility of a drug changes exponentially with change in pH as shown in the graphs.

3. In making injections, one dose of the drug is dissolved in around 2 ml of solvent. The pH of

blood is 7.4. When an injection at say pH 5 or 9 is given intra venously, due to change in pH the

drug will precipitate out and may block blood vessels. Hence buffers are included in injections.

4. The degree and extent of precipitation will depend on the ability of a formulation to resist pH

change when diluted.

5. Buffered injection, if given slowly by I.V route can avoid precipitation of solubilized drug.

6. Example: Amikacin sulfate I.V. injection has sodium citrate as buffer, pH is 3.5 to 5.5.

7. Example: Diazepam I.V. injection has benzoic acid and sodium benzoate as buffer, pH is 6 to 7.

8. Example: Phenytoin I.V. injection has sodium hydroxide, pH is 10 to 12.

Salt Formation:

1. Polar substances dissolve in polar solvents like water.

2. Ionized drug will have greater solubility in water than the unionized drug.

3. Hence many drugs are converted into salts to improve solubility.

4. Weakly acidic drugs are converted into salts by reacting with bases. Example: salicylic acid on

reacting with NaOH we get sodium salicylate.

5. Weakly basic drugs are converted into salts by reacting with acids. Example: Tetracycline is

converted into tetracycline HCl.

6. The below table gives solubility of drugs and their corresponding salts.

7. When a salt is added to water, the water molecules separate the ions and keep them in solution.

This is called hydration of ions.

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8. Some organic salts are very soluble in water, where as some are not.

9. Example: Naproxen is converted into potassium salt, sodium salt, magnesium salt, and calcium

salt. The potassium salt has the highest solubility, 200 times greater than calcium salt solubility.

S.NO Drug Solubility in 100 ml

water

Salt Form Solubility in 100

ml water

1 Salicylic acid 240 mg Sodium salicylate 124 g

2 Diclofenac 2.37 mg Diclofenac sodium 5 g

3 Tetracycline 170 mg Tetracycline HCl 1090 mg

Co Solvents:

1. A common method to increase the solubility of a non-polar drug is through the use of co solvents.

2. A co solvent system has water and a water miscible or partially miscible organic solvent.

3. Example of co solvents: Alcohol, glycerol, propylene glycol, PEG 400, di methyl sulfoxide

(DMS), di methyl acetamide (DMA).

4. Co solvents have some degree of hydrogen bond donating and or hydrogen bond accepting ability

as well as small hydrocarbon regions.

5. The resulting solution will have physical properties that are intermediate to that of the pure

organic solvent and water. Hence, the system is more favorable for non polar solutes.

6. The below table gives some solubility parameter of co solvents used in pharmacy. Octanol is also

added as a reference solvent. Drugs having large octanol water partition coefficients have poor

water solubility. By preparing cosolvent blends having solubility parameter close to octanol, we

can solubilize many organic drug molecules.

7. By mixing water and co solvent in different ratios, we can get a solvent blend with desired

solubility parameter. If solubility parameter of drug and solvent blend is nearly same, the drug

will have maximum solubility.

Solvent Solubility Parameter Solvent Solubility Parameter

Water 23.4 H Dimethyl sulfoxide 12 H

Glycerol 65 H Dimethyl acetamide 10.8 H

Propylene glycol 12.6 H Ethanol 12.7 H

PEG 400 11.3 H N – octanol 10.3 H

8. Example: Diazepam IV injection has 10 % ethanol and 40 % propylene glycol.

9. Example: Digoxin IV injection has 10 % ethanol and 40 % propylene glycol.

10. Example: Nitroglycerin IV injection has 70 % ethanol and 4.5 % propylene glycol.

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11. A mathematical relationship between solubility of drug and fraction of cosolvent is given below.

12. log Smix = log Sw + σ fc or log (Smix / Sw) = σ fc ---------- 9

Smix = solubility of drug in solvent mixture σ = solubilization slope

Sw = solubility of drug in water fc = fraction of co solvent

Co solvent Effect on Solubility of A Drug

0

1

2

3

4

5

6

7

0 0.2 0.4 0.6 0.8 1 1.2

Fraction of Co solvent

log

Sm

ix

Effect of Co solvents on Solubility of A drug

0

1

2

3

4

5

6

0 0.2 0.4 0.6 0.8 1 1.2

Fraction of co solvent

log

(S

mix

/ S

w)

Series1

Series2

Series 1: Alcohol

Series 2: Glycerol

13. A plot of log Smix versus fc will give a straight line with a slope equal to σ and a y intercept of

log Sw.

14. A plot of log (Smix / Sw) versus fc will give a straight line passing through the origin with a slope

equal to σ. This graph can be used for comparison purposes.

15. With many drugs, there is an exponential increase in solubility of a drug with co solvency.

16. In the above graph, the solubilization slope for alcohol is greater than that obtained for glycerol.

This indicates that alcohol is a better co solvent than glycerol for the drug.

17. The ability of a cosolvent to solubilize a drug depends on solute properties and cosolvent

properties.

18. The solubilization slope σ for a given cosolvent system depends on polarity of solute and

cosolvent and are related in the below equation. Polarity of drug is indicated by Ko/w and physical

properties of drug by s and t.

σ = s log Ko/w + t ---------- 10

Dependence of Solubilization on Solute Properties:

1. Isomers having same log Ko/w values have same degree of polarity and solubilization slopes will

be same for a given co solvent system. Differences in over all solubility are due to differences in

crystal contribution to solubility.

2. In case of non polar solutes (more non polar than the given co solvent), solubility increases

exponentially (log linear model) with use of co solvents. The degree of solubilization (σ)

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increases with log Ko/w values. S and t are constant in the below equation.

σ = s log Ko/w + t ---------- 10

3. The below graph shows the effect of propylene glycol on solubility of a series of

hydrocortisones. Higher esters are more non polar and solubilization slope is also more.

Effect of Solute Property on Solubilization

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.2 0.4 0.6 0.8 1 1.2

Fraction of Co Solvent

log

(S

mix

/Sw

)

Series1

Series2

Series3

Series4

Series5

Series6

Series 1: Hydrocortisone acetate Series 4: Hydrocortisone pentanoate

Series 2: Hydrocortisone propionate Series 5: Hydrocortisone hexanoate

Series 3: Hydrocortisone butyrate Series 6: Hydrocortisone heptanoate

4. For semi polar and polar solutes (polarities that are between water and the given co solvent),

there will be little increases in solubility due to co solvent. Water and co solvent should be mixed

in a definite ratio so that the solubility parameter of the solvent mixture matches with that of the

semi polar solute. The solute will have maximum solubility in this mixture.

5. The overall gain in solubility is less than a factor of five for semi polar solutes.

6. The solubility of polar solutes decrease with addition of co solvent. If a polar solute is insoluble

in water, it is the crystalline structure that is limiting the solubility. If the polar solute has an

acidic or basic group, adjust pH to improve solubility.

7. As hydrocarbon groups increase in the structure, the drug becomes more and more non polar and

solubility increases by 5 to 6 times due to co solvency.

8. As the drug is more non polar (more log Ko/w) greater the solubilization slope.

Dependence of Solubilization on Co solvent Properties:

1. When an organic co solvent is added to water, it reduces the hydrogen bonding between water

molecules and creates a suitable environment for dissolving non polar drug molecule.

2. If the co solvent is more non polar, the solvent mixture of water and co solvent also will be non

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polar and the solubilization of the non polar drug will be more.

3. Example: Benzocaine solubilization depended on co solvent non polar nature. The more non

polar the co solvent, greater the solubilization slope.

4. The below graph shows the effect of co solvent on solubilization of benzocaine.

Di methyl acetamide is the most non polar solvent and solubilization slope is highest.

Effect of Solvent Property on Solubilization

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.2 0.4 0.6 0.8 1 1.2

Fraction of Co Solvent

log

(S

mix

/Sw

)

Series1

Series2

Series3

Series4

Series5

Series6

Series 1: glycerol Series 4: PEG 200

Series 2: methanol Series 5: PEG 400

Series 3: ethanol Series 6: dimethyl acetamide

Multiple Co Solvents:

1. Multiple co solvents can be used for solubilization when we have limits for using a single co

solvent.

2. Example: In paracetamol elixir we cannot use more alcohol; hence we use glycerol, propylene

glycol as co solvents along with alcohol.

3. The effect of multiple co solvents on solubilization is given by the below equation.

log Smix = log Sw + σ1 fc1 + σ2 fc2+ σ3 fc3 + ……. --------- 11

4. The sub scripts 1, 2, 3, represent co solvents 1, 2 and 3.

5. Example: Solubilisation of spiranolactone was done using PEG 400 and propylene glycol as co

solvents.

6. Example: Solubilisation of spiranolactone was done using PEG 400 and glycerol as co solvents.

7. It was found that propylene glycol is a better solubilizing agent than glycerol in the above

example. The solubilization slope of propylene glycol was almost double that of glycerol.

8. From above equation it is assumed that, each co solvent acts independently in solubilization of

the drug, but in some cases, there will be synergistic effect also. The increase in solubility will be

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greater than that calculated theoretically from equation 11.

Summary of Co Solvency:

1. Co solvency is a powerful technique for solubilization of drugs.

2. With a given co solvent, degree of solubilization increases with log (octanol / water partition

coefficient) of the drug.

3. Solubility of a non polar drug increases exponentially with co solvents; hence there is a chance

for precipitation on dilution or on IV injection.

4. The limitation of this technique is the quantity of co solvent that we can use for solubilization.

Many co solvents have biological effects.

5. Example: Diazepam IV injection has 10 % ethanol and 40 % propylene glycol.

6. Example: Digoxin IV injection has 10 % ethanol and 40 % propylene glycol.

7. Example: Nitroglycerin IV injection has 70 % ethanol and 4.5 % propylene glycol

Surfactants:

1. The solubility of a drug in water can be increased by using surfactants. This technique is called

micellar solubilization.

2. Surfactants have two parts

a. Lipophillic tail / hydrophobic tail /non polar portion / oil loving portion.

b. Hydrophillic head / polar head /water loving portion.

3. Example: Potassium sterate, Tween 80, span 80, tween 20, span 20.

4. When surfactants are added to water, they occupy the interface. When there is no more space at

the surface, they form spherical aggregates in water. The tails are inside the sphere and the polar

heads are in contact with water as shown in the below figure.

5. Surfactants form micelles in water. They are spherical in shape. The inner core of the micelle is

lipophillic in nature and the outer surface is hydrophillic in nature. Drugs having poor solubility

in water are given accommodation inside the core of micelle. As a result the solubility of the drug

in water is increased. This technique is called micellar solubilization.

6. The concentration at which a surfactant starts forming micelles is called critical micellar

concentration (CMC). Below CMC, solubility of a non polar drug will not increase significantly.

Above CMC, there will be a significant increase in solubility of a drug.

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7. Example: CMC of tween 80 is 25 mg / 100 ml, above this concentration, micelles starts forming

and solubility of a non polar drug will increase significantly.

8. The below equation can be used to calculate the solubility of a drug by micellar solubilization.

ST = SW + K (Csurf – CMC) ----------- 12

ST = Total solubility of non polar drug, SW = Solubility of non polar drug in water

Csurf = Concentration of surfactant, CMC = Critical micellar concentration of surfactant

K = solubilization capacity.

If K value is more degree of solubilization will be more.

9. The graph for the above equation is given below. The solubility of the non polar drug does not

increase until CMC is crossed. Later, it increases linearly with concentration of tween 80.

10. Example: Tween 80 is used in multi vitamin syrups to solubilize vitamin A and D (fat soluble

vitamins).

11. Example: Chlordiazepoxide IV injecton has 4 % tween 80 as surfactant.

12. Taxol IV injection has 51 % cremophor EL as surfactant.

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Complexation:

1. Complexation is association of two or more molecules in a definite ratio to form a non bonded

substance.

2. Example: Caffeine is complexed with gentisic acid to form water insoluble complexes. It is used

in making chewable caffeine tablets. The taste is masked due to complexation.

3. The two type of complexation that are useful for increasing the solubility of drugs in aqueous

media are stacking complexation and inclusion complexation.

4. Stacking complexes are formed by the overlap of the planar regions of aromatic molecules, while

inclusion complexes are formed by the insertion of the non polar region of one molecule into the

cavity of another molecule (or group of molecules).

5. A complexation reaction is given below.

6. S + L → SL ,

7. S is drug / substrate and L is called ligand or complexing agent. In the above example, caffeine is

called substrate and gentisic acid is called ligand.

8. The equilibrium constant of a 1:1 complex, K 1:1, is given by the below equation. The equilibrium

constant is also known as the stability constant or the complexation constant. If this constant is

more, the complex is more stable.

K 1:1 = [ SL ] / [ S ] [ L ] ----------- 13

9. S is the concentration of the free solute, L is the concentration of the free ligand, and [SL] is the

concentration of the solute / ligand complex.

10. If it takes two ligand molecules to complex with a solute molecule the complexation constant is

defined by

K 1:2 = [ SL2 ] / [ S ] [ L ]2 ----------- 14

11. The general equation for solubilization by 1 : 1 complexation, is given by

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ST = Sw + [ K1:1 Sw / (1+ K1:1 Sw) ] Ltotal ----------- 15

12. The above equation is for a straight line. A plot of ST Vs L will give a straight line with slope of

σ complex = [ K1:1 Sw / (1+ K1:1 Sw) ] ------------ 16

13. The Y intercept is the aqueous solubility of the solute. Eq. 15 represents a linear increase in the

solubility of the solute with increasing ligand concentration.

14. Equation 16 can be rearranged to find out stability constant of complex.

K1:1 = σ complex / [Sw (1 - σ complex )] ---------- 17

15. This linear region will continue until the solution is saturated with complex also. Then, the total

solubility of the solute remains constant, as shown by the central part of the curve. Further

addition of the complexing agent can reduce solubility of solute as shown in the last part of the

curve.

16. The solubilization curve for a solute molecule that complexes with two ligand molecules is more

complicated than those shown in the above graph.

17. If the complexation constant for a second ligand is significantly lower than the first, a 1 : 1

complex will be formed at lower ligand concentration. It will then combine with a second ligand

to produce a 2 : 1 complex.

18. If 2:1 complex is more soluble than the 1 : 1 complex, the solubilization curve will have two

distinct slopes. If each ligand is equally capable of complexing with the solute, they will complex

simultaneously to produce a convex up solubilization curve.

19. The factors that determine the degree of solubilization of the solute are the complexation constant

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and the solubilities of the solute, the ligand, and the complex. Hence, the most useful ligands for

solubilization in aqueous media should be highly water soluble, and produce soluble complexes.

Self-Association and Stacking Complexation:

1. Non polar molecules are squeezed out of water by the strong hydrogen bond interactions present

in water.

2. These non polar molecules want to minimize their contact with water. This can be done by

stacking one above another like one playing card above another.

3. This type of self association is seen with molecules having large planar non polar regions.

4. Some examples of substances undergoing stacking interaction are shown below.

Inclusion Complexes:

1. In this type of complex, a non polar molecule (guest) is given accommodation in the non polar

cavity of another molecule (host).

2. When the guest molecule enters into the host cavity, the contact area between water and the non

polar regions of both is reduced.

3. This complexation is also due to squeezing out of guest n host by water molecules.

4. This complexation will be successful, if there is a snug fit (tight fit) between guest and host

molecule.

5. The most commonly used host molecules are cyclo dextrin’s. These are cyclic oligomers of

glucose having cavities to accommodate non polar portions of drug molecules.

6. The naturally occurring cyclo dextrins have 6, 7, 8, gluco pyranose units and are known as α, β, γ

cyclo dextrins.

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7. α cyclo dextrin has six gluco pyranose units and can give accommodation to small alkyl groups in

their cavity.

8. β cyclo dextrin has seven gluco pyranose units and can give accommodation to molecules having

a single aromatic ring.

9. γ cyclo dextrin has eight gluco pyranose units and can give accommodation to large hydrocarbon

molecules.

10. These cyclo dextrins have limited solubility in water and are modified to improve their solubility.

Example: Hydroxy propyl beta cyclo dextrin.

11. The degree of solubilization of a non polar drug molecule depends on several properties.

a. The drug should have a significant non polar portion to be squeezed out of water and into

cyclo dextrin cavity.

b. Whole guest molecule or a significant part of fixed molecule must fit into the cavity.

c. The tight fit and inter molecular interactions and between guest and host will be responsible

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for stability of the complex.

12. Example: Testosterone solubility has been increased by 2020 times with 0.1 M SBE – β CD

( sulfo butyl ether beta cyclo dextrin).

13. Example: Chloramphenicol solubility has been increased by 8 times with 0.1 M SBE – β CD

( sulfo butyl ether beta cyclo dextrin).

14. Example: Dexamethasone solubility has been increased by 208 times with 0.1 M SBE – β CD

( sulfo butyl ether beta cyclo dextrin).

15. Seo et.al carried out complexation studies with spiranolactone and the three cyclo dextrins (α, β,

and γ cyclo dextrin) and found that stability constant and solubility enhancement depended on

type of cyclo dextrin. The graph is shown below.

From the slopes of the lines, it can be seen that

stability constant is maximum with gamma

cyclo dextrin. Gamma cyclo dextrin has a large

cavity to give accommodation for the big

spiranolactone molecule. Solubility

enhancement is more with gamma cyclo dextrin

when compared with other Cyclodextrins.

Combination of pH and Complexation:

1. The effect of pH on solubilization by complexation will depend on the drug and ligand.

2. Tinwalla et al. found that combination of pH and complexation can be a powerful tool for

solubilization. They increased solubility of thiazolo benz imidazole (TBI) by use of pH and

complexation with hydroxyl propyl – β cyclodextrin.

3. If the ligand or drug ionizes with change in pH, stability constant will decrease. The stability

constant of the ionized complex of TBI was less by 13 times when compared with the neutral

complex. However the solubility increased by 3 times when compared with the neutral complex.

4. Hence we can try pH adjustment and complexation for solubilization of lipophillic drugs.

Dr. Eswar Gupta Maddi, M.Pharm., Ph.D.,Professor and Head, Dept. of Pharmaceutics, Sir C.R. Reddy College of Pharmaceutical Sciences, Eluru, West Godavari district, A.P, India 534007. Cell: +91 988 55 237 60, email: meguptas@yahoo.com, meguptas@gmail,com

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