There are two systems of weights and measures:
A. The imperial system
B. The metric system
IMPERIAL SYSTEM
It is an old system of weights and measures.
Measurements of weights in imperial system
Weight is a measure of the gravitational force acting on a body and is directly proportional to its mass.
The imperial systems are of two types: (a) Avoirdupois system and (b) Apothecaries system
(a) Avoirdupois system
In this system pound (lb) is taken as the standard of weight (mass).
TABLE:
1 pound avoir (lb)
= 16 oz avoir
oz is pronounced as ounce.
1 pound avoir (lb)
= 7000 grains (gr)
(b) Apothecary or Troy system
In this system grain (gr) is taken as the standard of weight (mass).
TABLE:
1 pound apoth (lb)
= 12 ounces ( )
1 pound apoth (lb) = 5760 grains (gr)
1 ounce ( )
= 8 drachms ( )
1 drachm ( )
= 3 scruples (')
1 scruple (')
= 20 grains (gr)
Measurements of volumes.
TABLE:
1 gallon (c)
= 4 quart
1 quart
= 2 pint (o)
1 pint (o)
= 20 fluid ounce
1 fluid ounce
= 8 fluid drachm
1 fluid drachm
= 3 fluid scruple
1 fluid scruple
= 20 minims
Exercise:
Convert (i) quart to minim
1 quart = 2 pint
= 2x (20 fluid ounce)
= 2x20x (8 fluid drachm)
= 2x20x8x (3 fluid scruple)
= 2x20x8x3x(20minims)
= 19200 minims
(ii) pint to fluid ounce, (iii) fluid ounce to minim, fluid drachm = minim
THE METRIC SYSTEM
‘Kilogram’ is taken as the standard weight (mass).
1 kilogram (kg)
= 1000 grams (g)
Kilo = 1000 Greek word
1 hectogram (hg)
= 100 grams (g)
Hecto = 100 Greek word
1 dekagram (dg)
= 10 grams (g)
Deka = 10 Greek word
1 gram (g)
1 gram (g)
1 decigram (dcg)
1/10 gram (g)
Deci = 1/10 Latin word
1 centigram (cg)
1/100 gram (g)
Centi = 1/100 Latin word
1 milligram (mg)
1/1000 gram (g)
Milli = 1/1000 Latin word
1 microgram (mg, mcg)
10–6 gram (g)
Micro = 10–6.
1 nanogram (ng)
10–9 gram (g)
Nano = 10–9.
Measurement of volume
‘Litre’ is taken as the standard of volume.
1 liter (L, lit)
1000ml
1 microliter (ml)
1/1000 ml
CONVERSION TABLE
Domestic measures
Metric System
Imperial system
1 drop
0.06ml
1 minim
1 teaspoonful
5 ml
1 fluid drachms
1 desert spoonful
8 ml
2 fluid drachms
1 tablespoonful
15 ml
4 fluid drchms
1 wine-glassful
60 ml
2 fluid ounces
1 teacupful
120 ml
4 fluid ounces
1 tumblerful
240 ml
8 fluid ounce
Weight measure conversion table
1 kilogram
= 2.2 pounds (lb)
1 ounce apoth.
= 30 g
1 pound avoir.
= 450 g
1 grain
= 65 mg
POSOLOGY
POSOLOGY is derived from the Greek word posos meaning how much and logos meaning science. So posology is the branch of medicine dealing with doses.
The optimum dose of a drug varies from patient to patient. The following are some of the factors that influence the dose of a drug.
1. Age: Human beings can be categorized into the following age groups:
1. Neonate: From birth up to 30days.
2. Infant: Up to 1 year age
3. Child in between 1 to 4 years
4. Child in between 5 to 12 years.
5. Adult
6. Geriatric (elderly) patients
In children the enzyme systems in the liver and renal excretion remain less developed. So all the dose should be less than that of an adult. In elderly patients the renal functions decline. Metabolism rate in the liver also decreases. Drug absorption from the intestine becomes slower in elderly patients. So in geriatric patients the dose is less and should be judiciously administered.
2. Sex: Special care should be taken while administering any drug to a women during menstruation, pregnancy and lactation. Strong purgatives should not be given in menstruation and pregnancy. Antimalarials, ergot alkaloids should not be taken during pregnancy to avoid deformation of foetus. Antihistaminic and sedative drugs are not taken during breast feeding because these drugs are secreted in the milk and the child may consume them.
3. Body size: It influences the concentration of drug in the body. The average adult dose is calculated for a person with 70kg body weight (BW). For exceptionally obese (fat) or lean (thin) patient the dose may be calculated on body weight basis.
Another method of dose calculation is according to the body surface area (BSA). This method is more accurate than the body weight method.
The body surface area (BSA) of an individual can be obtained from the following formula:
BSA (m2) = BW(kg)0.425 x Height (cm)0.725 x 0.007184
4. Route of administration
In case of intravenous injection the total drugs reaches immediately to the systemic circulation hence the dose is less in i.v. injection than through oral route or any other route.
5. Time of administration
The drugs are most quickly absorbed from empty stomach. The presence of food in the stomach delays the absorption of drugs. Hence a potent drug is given before meal. An irritant drug is given after meal so that the drug is diluted with food and thus produce less irritation.
6. Environmental factors
Stimulant types of drug are taken at day time and sedative types of drugs are taken at night. So the dose of a sedative required in day time will be much higher than at night.
Alcohol is better tolerated in winter than in summer.
7. Psychological state
Psychological state of mind can affect the response of a drug, e.g. a nervous and anxious patient requires more general anaesthetics. Placebo is an inert substance that does not contain any drug. Commonly used placebos are lactose tablets and distilled water injections. Some time patients often get some psychological effects from this placebo. Placebos are more often used in clinical trials of drugs.
8. Pathological states (i.e. Presence of disease)
Several diseases may affect the dose of drugs:
In gastrointestinal disease like achlorhydria (reduced secretion of HCl acid in the stomach) the absorption of aspirin decreases.
In liver disease (like liver cirrhosis) metabolism of some drugs (like morphine, pentobarbitone etc.) decreases.
In kidney diseases excretion of drugs (like aminoglycosides, digoxin, phenobarbitone) are reduced, so less dose of the drugs should be administered.
9. Accumulation
Any drug will accumulate in the body if the rate of absorption is more than the rate of elimination. Slowly eliminated drugs are often accumulated in the body and often causes toxicity e.g. prolonged use of chloroquin causes damage to retina.
10. Drug interactions
Simultaneous administration of two drugs may result in same or increased or decrease effects.
Drug administration with dose
Pharmacological effect
Drug A
Effect A
Drug B
Effect B
Drug A + Drug B
Effect AB
Relationship
Name of the effect
Examples
Effect AB = Effect A + Effect B
Additive effect
Aspirin + Paracetamol
Effect AB > Effect A + Effect B
Synergistic (or potentiation)
Sulfamethaxazole + Trimethoprim
Effect AB < Effect A + Effect B
Antagonism
Histamine + Adrenaline
11. Idiosyncrasy
This an exceptional response to a drug in few individual patients. For example, in some patients, aspirin may cause asthma, penicillin causes irritating rashes on the skin etc.
12. Genetic diseases
Some patients may have genetic defects. They lack some enzymes. In those cases some drugs are contraindicated.
e.g. Patients lacking Glucose-6-phosphate dehydrogenase enzyme should not be given primaquin (an antimalarial drug) because it will cause hemolysis.
13. Tolerance
Some time higher dose of a drug is required to produce a given response (previously less dose was required).
Natural Tolerance: Some races are inherently less sensitive to some drugs, e.g. rabbits and black race (Africans) are more tolerant to atropine.
Acquired Tolerance: By repeated use of a drug in an individual for a long time require larger dose to produce the same effect that was obtained with normal dose previously.
Cross tolerance: It is the development of tolerance to pharmacologically related drugs e.g. alcoholics are relatively more tolerant to sedative drugs.
Tachyphylaxis: (Tachy = fast, phylaxis = protection) is rapid development of tolerance. When doses of a drug is repeated in quick succession an reduction in response occurs – this is called tachyphylaxis. This is usually seen in ephedrine, nicotine.
Drug resistance: It refers to tolerance of microorganisms to inhibitory action of antimicrobials e.g. Staphylococci to penicillin.
CALCULATIONS OF DOSES FOR CHILDREN
A number of methods have been used to relate doses for children to their ages.
1. Dose proportionate to age
Young’s formula: This formula is used for children having age below 12 years.
Dilling’s formula: This formula is used for children having age from 4 to 20 years. This formula is better because it is easier to calculate the dose.
Cowling’s formula:
Freud’s formula: For less than 12 years of age
2. Doses proportionate to body weight
Clark’s formula:
3. Doses proportionate to body surface area (BSA)
TABLE: Calculation of child doses
Age
Weight (kg)
Height (cm)
BSA (m2)
Fraction of adult dose
Young’s rule
Clark’s Rule
BSA method
Birth
3 mos
6 mos
1 yr
2 yrs
3 yrs
4 yrs
5 yrs
6 yrs
7 yrs
8 yrs
9 yrs
10 yrs
11 yrs
12 yrs
3.5
5.7
7.5
9.9
12.5
14.5
16.5
19.1
21.5
24.2
26.9
29.5
32.3
35.5
39.1
50.5
59.9
65.8
74.7
86.9
96.0
103.4
110.5
116.8
123.2
129.0
134.1
139.4
144.5
150.9
0.21
0.29
0.35
0.44
0.54
0.61
0.68
0.76
0.84
0.91
0.98
1.04
1.12
1.20
1.28
–
0.02
0.04
0.08
0.14
0.20
0.25
0.29
0.33
0.37
0.40
0.43
0.45
0.48
0.60
0.05
0.08
0.11
0.15
0.18
0.21
0.24
0.28
0.32
0.35
0.39
0.43
0.47
0.52
0.57
0.12
0.17
0.20
0.25
0.31
0.35
0.39
0.44
0.49
0.53
0.57
0.60
0.65
0.69
0.74
Exerxise: What will be the dose for a child of 6 years if the adult dose is 500mg.
REDUCING AND ENLARGING FORMULAE (RECIPE)
In order to prepare any pharmaceutical product, it is necessary to make it from a master formula or official formula. This master formula may be scaled down or scaled up depending on the requirement.
Rules for conversion of the formula
1. Determine the total weight or volume of the whole preparation.
2. Calculate the ratio of . This is called conversion factor.
3. Multiply the conversion factor with the quantity of each ingredient. The unit should be unchanged.
Example of reducing the recipe
The master formula: Give the working formula for 100ml preparation.
Ingredient
Quantity required
Drug X
Sucrose
Purified water q.s.
120g
480g
1000ml
The total volume of the preparation is 1000ml. Required volume of the preparation is 100ml.
So the conversion factor is
The reduced formula
Ingredient
Quantity required for 1000ml
Conversion factor
Quantity required for 100ml
Drug X
Sucrose
Purified water q.s.
120g
480g
1000ml
100/1000 = 0.1
12.0g
48.0g
100ml
Example of enlarging the recipe
The master formula: Give the working formula for 2.5 L
Ingredient
Quantity required
Liquid P
Solid A
Liquid R
Liquid S
Purified water q.s.
35ml
9g
2.5ml
20ml
100ml
Total volume of the preparation is 100ml. Required volume of the preparation is 2.5 L i.e. 2500ml.
So the conversion factor is
The enlarged formula
Ingredient
Quantity required for 1000ml
Conversion factor
Quantity required for 100ml
Liquid P
Solid A
Liquid R
Liquid S
Purified water q.s.
35ml
9g
2.5ml
20ml
100ml
2500/100=25
875ml
225g
62.5ml
500ml
2500ml
Exercise: Calculate the mount of ingredients required for preparing 30g of ointment.
Ingredient
Quantity required for 1000g
Conversion factor
Quantity required for 30g
Wool fat
Hard Paraffin
Cetostearyl alcohol
White soft paraffin
50g
50g
50g
850g
30/1000 = 0.03
1.5g
1.5g
1.5g
25.5g
Total = 1000g
PERCENTAGE SOLUTIONS
The concentration of a substance can be expressed in the following three types of percentages:
1. Weight in volume (w/v) : Required to express concentration of a solid in liquid.
2. Weight in weight (w/w) : Required to express concentration of a solid in solid mixture.
3. Volume in volume (v/v) : Required to express concentration of a liquid in another liquid.
Weight in volume (w/v)
In this case the general formula for 1%(w/v) is:
Solute 1part by weight
Solvent upto 100 parts by volume
The formula is actually:
Solute 1 g
Solvent upto 100 ml
Exercise1: Calculate the quantity of sodium chloride required for 500ml of 0.9% solution.
Ans: 0.9%w/v solution of sodium chloride =
So 500ml solution will contain sodium chloride
Exercise2: Send 100ml of a solution of potassium permanganate of which one part diluted with seven parts of water makes a 1 in 8000 solution.
Ans. The planning of calculation is as follows:
Original solution
Solution of potassium permanganate,
x % w/v, 100ml
Dilution of the solution
Solution, x % w/v = 1ml
Water = 7ml
Volume of solution = 8ml
Final solution after dilution
Potassium permanganate = 1g
Volume of solution = 8000ml
So, we have to calculate x. Let us start from final solution.
Concentration of KMnO4 is the final solution = = 0.0125 %w/v
Method-1
Let us restructure the problem:
1 ml of x% w/v solution is diluted to a solution of 0.0125%w/v and the final volume is 8ml.
V1 = 1ml V2 = 8ml
S1 = x%w/v S2 = 0.0125%w/v
V1 x S1 = V2 x S2
Or, 1ml x X% = 8ml x 0.0125%
Or,
Or, X% = 0.1%
Or, X = 0.1
Method-2
Concentration of initial solution = ?
Concentration of diluted solution = 0.0125%(w/v)
1 ml diluted to 8ml, so dilution factor = 8, i.e. the solution is diluted 8 times
Concentration of initial solution = Concentration of diluted solution x 8 = 0.0125% w/v x 8 = 0.1%w/v
Ans. A 0.1%w/v potassium permanganate solution is to be prepared.
Exercise 3: Send 250ml of 4 percent potassium permanganate solution and label with directions for preparing 1 liter quantities of a 1 in 2500 solution.
Ans. The planning of calculation is as follows:
Original solution
Solution of potassium permanganate,
4 % w/v, 250ml
Dilution of the solution
Solution, 4 % w/v = 1ml
Water = ?
Final solution after dilution
Potassium permanganate = 1g
Volume of solution = 2500ml
Now do it yourself. Do it by Method-2.
Ans: 100 times dilution i.e. 1 ml is diluted with 99ml water to obtain 100ml solution.
Weight in weight (w/w)
In this case the general formula for 1%(w/w) is:
Solute 1part by weight
Solvent upto 100 parts by weight
The formula is actually:
Solute 1 g
Solvent up to 100 g
Problem: Prepare 100ml Phenol Glycerin BPC. It contains 16%w/w phenol in glycerol. Sp.gr. of glycerol = 1.26
Let us assume that phenol is not increasing the volume of the solution.
So the final solution: Volume = 100ml
Volume of glycerol = 100ml
Weight of glycerol = 100ml x 1.26 g/ml = 100 x 1.26 g = 126g
So the working formula will be:
Ingredient
Quantity for 100g
Quantity required for 100ml
Glycerol
Phenol
84g
16g
126g
= 24g
Volume in volume (v/v)
In this case the general formula for 1%(w/w) is:
Solute 1part by volume
Solvent upto 100 parts by volume
The formula is actually:
Solute 1 ml
Solvent upto 100 ml
Problem: Prepare 600ml of 60%v/v alcohol from 95% v/v alcohol.
In this problem: V1 = ? S1 = 95% V2 = 600ml S2 = 60%
V1 x S1 = V2 x S2 or, V1 = = 379ml
Ans: 379 ml of 95% alcohol is diluted to 600ml to obtain 60% alcohol.
CALCULATION BY ALLIGATION METHOD
This types of calculation involves the mixing of two similar preparations, but of different strengths, to produce a preparation of intermediate strength. The name is derived from the Latin alligatio, meaning the act of attaching and hence referes to the lines drawn during calculation to bind quantities together.
Method:
Example:
Prepare 600ml of 60%v/v alcohol from 95% v/v alcohol.
Higher concentration = 95%
Required concentration = 60%
Lower concentration = 0% (i.e. water)
So from alligation method it is obtained:
Volume of 60% alcohol solution = 600ml
\the volume of 95% alcohol required = = 379ml
PROOF SPIRITS
For excise (tax) purpose, the strength of alcohol in indicated by degrees proof.
The US System: Proof spirit is 50% alcohol by volume (or 42.49% by weight).
The British / Indian system: Proof spirit is 57.1% ethanol by volume (or 48.24% by weight.
Definition: Proof spirit is that mixture of alcohol and water, which at 510F weighs 12/13th of an equal volume of water.
[N.B. Density of proof spirit = 12/13 of density of water at 510F = 0.923 g/ml]
This means that any alcoholic solution that contains 57.1%v/v alcohol is a proof spirit and is said to be 100 proof.
100 degree proof alcohol = 57.1% v/v alcohol
If the strength of the alcohol is above 57.1%v/v alcohol then the solution is called “over proof”.
If the strength of the alcohol is below 57.1%v/v alcohol then the solution is called “under proof”.
In India, the excise duty is calculated in terms of Rupees per litre of proof alcohol. So any strength of alcohol is required to be converted to degree proof . We shall follow the British System
Conversion of strength of alcohol from %v/v to degrees proof as per Indian system.
Strength of alcohol = x 100
Conversion of strength of alcohol from degrees proof to %v/v as per Indian system.
Strength of alcohol in %v/v =
Example 1:Find the strength of 95%v/v alcohol in terms of proof spirit.
Strength of alcohol = x 100 = 166.34 degree proof = (166.34-100) degrees over proof = 66.34 0 op
Example 2:Find the strength of 20%v/v alcohol in terms of proof spirit.
Strength of alcohol = x 100 = 35.03 degree proof = (100-35.03) degrees under proof = 64.97 0 up
Example 3:Calculate the real strength of 300op and 400up.
300op = (100 + 30) = 130 deg proof Therefore the strength of alcohol = = 74.23%v/v
400op = (100 – 40) = 60 deg proof Therefore the strength of alcohol = = 34.26%v/v
Example 4:How many proof gallons are contained in 5 gallon of 70%v/v alcohol?
1 proof gallon = 1 gallon proof alcohol = 1 gallon of 100 degrees proof alcohol
70% v/v alcohol = x 100 degrees proof alcohol
= 122.59 degrees proof alcohol
= proof alcohol = 1.226 proof alcohol
5 gallons 70%v/v alcohol = 5 gallons of 1.226 proof alcohol
= 6.13 proof gallon
pH AND BUFFER SOLUTIONS
A proton binds with a molecule of water to produce a hydronium ion, i.e. H2O + H+ = H3O+.
Mathematically the pH of a solution is defined as the negative logarithm of hydrogen ion (more appropriately hydronium H3O+ ) concentration in molarity.
pH = – log [H3O+]
Buffer / buffer solution / buffered solution refers to the ability of an aqueous solution to resist a change of pH on adding acid or alkali, or on dilution with a solvent.
N.B. Distilled water has very little buffer action, hence carbon dioxide of air, when equilibrated with distilled water (pH = 7.0), the pH of the water changes to 5.7.
A solution will show buffer action if a conjugate acid-base pair is present in the solution.
e.g.
The dissociation constant, Ka =
Taking logarithm of both hand sides we get,
log Ka = log [CH3COO–] + log [H3O+] – log [CH3COOH]
Multiplying –1 with both hand sides yield:
– log Ka = – log [H3O+] + log [CH3COOH] – log [CH3COO–]
or, pKa = pH + log [CH3COOH] – log [CH3COO–]
or, pH = pKa – log [CH3COOH] + log [CH3COO–]
or, pH = pKa +
or, pH = pKa + This equation is called Henderson-Hasselbalch equation.
This ratio of and Ka determines the pH of the solution. For a certain weak acid or base Ka is constant, so if the ratio of concentrations of the [base] / [acid] is changed the pH of the buffer solution can be changed.
This equation can be used in the following buffer systems:
Name of the buffer system
Conjugate acid
Conjugate base
Acetic acid – Sodium acetate buffer
Acetic acid (CH3COOH)
Acetate ion (CH3COO– )
Ammonia – Ammonium chloride buffer
Ammonium ion (NH4+)
Ammonia (NH3)
Monosodium phosphate – Disodium phosphate
Monosodium phosphate
(NaH2PO4)
Disodium phosphate
(Na2HPO4)
Phenobarbital – Sodium phenobarbital
Phenobarbital
Sodium phenobarbital
Use of Henderson – Hasselbalch equation
1. The pH of a buffer solution can be calculated if the pKa, concentration of the base and acid are known.
2. During preparation of a buffer solution the ratio of the concentration of the conjugate acid and base pair can be calculated.
3. To calculate the buffer capacity of a buffer soltution.
Problem-1: What will be the pH of a solution containing acetic acid and sodium acetate, each in 0.1M concentration. Ka of acetic acid is 1.8 x 10–5 at 250C.
Ans: pKa = – log Ka = – log 1.8 x 10–5. = – (log 1.8 + log 10–5) = – (0.26 – 5) = – (–4.74) = 4.74
Concentration of acid = [acid] = [CH3COOH] = 0.1M
Concentration of base = [base] = [CH3COO –] = 0.1 M
From Hender- Hasselbalch equation we get
pH = pKa + = 4.74 + = 4.74 + log 1 = 4.74 + 0 = 4.74 Ans.
Problem-2: An acetic acid- acetate buffer is to be prepared having pH 4.5. What will be the ratio of the molar concentration of the acid base pair. Given pKa of acetic acid = 4.74.
Ans: Using Henderson – Hasselbalch equation we get:
pH = pKa + or, pH – pKa =
or, =antilog (pH – pKa) = 10 (pH – pKa) = 10 (4.5 – 4.74) = 10 –0.24 = 0.575
The answer is [sodium acetate] : [acetic acid] = 0.575 : 1BUFFER CAPACITY
The ability of a buffer solution to resist changes in pH upon addition of acid or alkali is measured in terms of buffer capacity of the solution.
Van Slyke has defined the buffer capacity as follows:
The amount (gm-equivalent) of strong acid or strong base,
required to be added to a solution to change its pH by 1 unit.
In mathematical form: Buffer capacity of a solution =
Problem-3: (a) What is the change of pH on adding 0.01mol of NaOH to 1 L of 0.10 M acetic acid? (b)Calculate the buffer capacity of the acetic solution. Ka = 1.75 x 10–4.
Ans: (a) Calculation of pH of 0.1 M solution of acetic acid
[H3O+] = = 4.18 x 10–3.
Therefore pH = – log (4.18 x 10–3 ) = – (–2.38) = 2.38
(b) On adding 0.01moles of NaOH, 0.01 mol of acetic acid will be converted to form 0.01 mol of acetic acid.
So after addition of NaOH [CH3COO–] = 0.01mol/ L = 0.01M
[CH3COOH] = (0.10mol – 0.01mol)/L = 0.09 mol / L = 0.09 M
Applying Henderson – Hasselbalch equation to calculate the pH of the final solution we get:
pH = pKa + = 4.76 + = 4.76 + (–0.954) = 3.81
Therefore the change in pH after addition of NaOH = final pH – initial pH = 3.81 – 2.38 = 1.43
So, from definition the
Buffer capacity of the solution = = 0.007 Ans.
Problem-4: (a) What is the change of pH on adding 0.01mol of NaOH to 1 L of buffer solution of 0.10 M acetic acid 0.1M of sodium acetate? (b)Calculate the buffer capacity of the solution. Ka = 1.75 x 10–4.
Ans: (a) The pH of the buffer solution before addition of NaOH is
[base] = [CH3COO–] = 0.1M
[acid] = [CH3COOH] = 0.1 M
pH = pKa + = 4.76 + = 4.76 + log (1) = 4.76 + 0 = 4.76
(b) On adding 0.01mol of NaOH per litre to this buffer solution 0.01mol aicd will be converted to base:
[base] = [CH3COO–] = (0.10mol + 0.01mol) / L = 0.11mol / L = 0.11 M
[acid] = [CH3COOH] = (0.10mol – 0.01mol) / L = 0.09 mol/L = 0.09 M
pH = pKa + = 4.76 + = 4.76 + log (1.22) = 4.76 + 0.09 = 4.85
Therefore the change in pH after addition of NaOH = final pH – initial pH = 4.85 – 4.76 = 0.09
So, from definition the
Buffer capacity of the solution = = 0.111 Ans.
So, this buffer solution has greater buffer capacity (0.111) than the solution in problem-3 (0.007).
ISOTONIC SOLUTIONS
Osmosis: If a solution is placed in contact with a semipermeable membrane the movement of the solvent molecules through the membrane is called osmosis.
An ideal semipermeable membrane only lets the solvent molecules to pass through it but not the solute molecules. The biological membranes are not ideal semipermeable membranes. They are selectively permeable; they give passage to some solutes while stop the passage of others. In case of biological membranes another term tonicity is used.
Isotonicity: A solution is isotonic with a living cell if there is no net gain or loss of water by the cell, when it is in contact with this solution.
If a living cell is kept in contact with a solution and there is no loss or gain of water by the cell then the solution is said to be isotonic with the cell.
· It is found that the osmotic pressure of 0.9%w/v NaCl solution is same as blood plasma. So 0.9%w/v NaCl solution is isotonic with plasma.
Tonicity– A. Isotonic – When a solution has same osmotic pressure as that of 0.9%w/v NaCl solution.
B. Paratonic – Not isotonic
(a) Hypotonic – The osmotic pressure of the solution is higher than 0.9%w/v NaCl solution
(b) Hypertonic – The osmotic pressure of the solution is lower than 0.9%w/v NaCl solution
Test of tonicity
A red blood corpuscle is placed in a solution and after some time it is viewed under microscope.
Observation
Conclusion
Mechanism
The shape and size of the cell remained unchanged
The solution is isotonic
Osmotic pressure of the cell fluid and the solution are same. No movement of water occurs across the cell membrane.
The size of the cell increased and may burst.
The solution is hypotonic.
Osmotic pressure of the cell fluid is more than the solution. Water molecules moved from the solution to the interior of the cell, so the cell swelled.
The size of the cell is reduced or shrinked.
The solution is hypertonic.
Osmotic pressure of the cell fluid is less than the solution outside. Water molecule moved from the interior of the cell to the solution.
N.B. If the red blood cell bursts then the hemoglobin comes out of the cell and the plasma become red in color. This phenomenon is called haemolysis.
Importance of adjustment of tonicity in pharmaceutical dosage forms
1. Solution for intravenous injection: The injection must be isotonic with plasma, otherwise the red blood corpuscle may be haemolysed.
2. Solution for subcutaneous injection: Isotonicity is required but not essential, because the solution is coming in contact with fatty tissue and not in contact with blood.
3. Solution for intramuscular injection: The aqueous solution may be slightly hypertonic. This will draw water from the adjoining tissue and increase the absorption of the drug.
4. Solution for intracutaneous injection: Diagnostic preparations must be isotonic, because a paratonic solution may cause a false reaction.
5. Solutions for intrathecal injection: Intrathecal injections are introduced in the cavities of brain and spinal chord. It mixes with the cerebrospinal fluid (CSF). The volume of CSF is only 60 to 80ml. So a small volume of paratonic injection may change the osmotic pressure of the CSF, which may lead to vomiting and other side effects.
6. Solutions for nasal drops: Aqueous solutions applied within the nostril may produce irritation if it is paratonic. So nasal drops must be isotonic with plasma.
7. Solutions for ophthalmic use: Only one or two drops of ophthalmic solutions are generally used. So it is not essential for eyedrops to be isotonic. Slight paratonicity will not produce great irritation because the eyedrops will be diluted with the lachrymal fluid.
Calculations for adjustment of tonicity
N.B. It is difficult and time consuming to determine the osmotic pressure of a solution. So some indirect methods are adopted to compare between two isotonic solutions. Two solutions will produce same osmotic pressure if both contain the same numbers of ultimate units. These units may be as follows:
1. These units may be molecules in case of substances those do not ionize.
2. These units may be ions in case of substances those ionize.
3. These units may be both ions and unionized molecules in case of weak electrolytes.
Some physical properties of these solutions depend on this number (or, collection) of units, such as osmotic pressure, freezing point depression (DTf), vapor pressure etc. – these physical properties are called colligative properties of the solutions.
Since these colligative properties are inter-dependent, so osmotic pressures of two solutions can be compared from their colligative properties like freezing point depression.
Tonicity of a solution can be adjusted by the following methods:
1. Freezing point depression method (DTf)
2. Sodium chloride equivalent method (E)
3. Isotonic solution V-Value method
1. Freezing Point Depression Method
Theory: Freezing point of pure water is 00C. When any impurities are there (like salt, drug etc.) the water freezes at some lower temperatures (like –0.180C). In case of a solution the solute units reduces the freezing point of water.
So the freezing point depression, DTf = Freezing point of pure water – Freezing point of the solution
This DTf is proportional to the number of units of solutes present in the solution.
DTf is also proportional to the osmotic pressure of the solution.
Now while preparing an injection or ophthalmic solution the drug is given in a certain percentage (i.e. %w/v)
This solution generally is hypotonic. In this solution some inert solute (like NaCl or dextrose) is dissolved to raise the osmotic pressure up to the osmotic pressure of serum (or plasma).
This problem is solved in three steps:
Step-I
Identify a reference solution and the associated tonicity parameter (e.g. freezing point depression, NaCl equivalent value E, or V-value etc.)
Step-II
Determine the contribution of the drug(s) and additive(s) to the total tonicity
Step-III
Determine the amount of sodium chloride (or dextrose) needed by subtracting the contribution of the original solution from the reference solution.
The freezing point of plasma =–0.520C
The freezing point depression of plasma = Freezing point of pure water – Freezing point of plasma
= 00C – (–0.520C)
= 0.520C
The freezing point depression of sodium chloride = 0.520C
TABLEs (See Remington p.622) are available where the name of drug and the “D values” are given.
e.g.
Drug
D- values of the following solutions
0.5%
1%
2%
3%
5%
Iso-osmotic concentration
Dexamethasone sodium phosphate
0.050
0.095
0.180
0.260
0.410
0.52
Naphazoline hydrochloride
0.140
0.52
Sodium chloride
0.576
1.73
2.88
0.52
Dextrose
0.091
0.28
0.46
0.52
Example 1:
Rx Dexamethasone sodium phosphate 0.1%w/v
Purified water q.s. 30ml
Prepare an isotonic solution
Step-I: Reference solution: 0.9%w/v NaCl solution
DTf = 0.520C
Step-II: Contribution of the drug to the freezing point depression of the solution
0.5%w/v dexamethasone sodium phosphate contributes a DTf of 0.0500C
\ 0.1%w/v dexamethasone sodium phosphate contributes a DTf of 0.0500C = 0.0100C
Step-III: Contribution of reference solution – Contribution of actual solution
= 0.520C – 0.010C = 0.510C
0.520C of DTf is contributed by 0.9%w/v NaCl solution
\ 0.510C of DTf is contributed by NaCl solution = 0.883% w/v NaCl solution
So total quantity of NaCl required = 0.883% x 30ml = = 0.265 g NaCl
Ans: To a 30ml solution of 0.1%w/v dexamethasone sodium phosphate, sodium chloride required is 0.265g to produce an isotonic solution.N.B. Let us summarize the whole calculation
Ingredient
DTf.
Dexamethasone sodium phosphate
0.1%w/v
0.010C
Sodium chloride
0.883%w/v
0.510C
Total
0.520C
2. Sodium Chloride Equivalent Method (E)
A sodium chloride equivalent, “E value” is defined as the weight of sodium chloride that will produce the same osmotic effect as 1g of the drug.
TABLEs (See Remington p.622) are available where the name of drug and the “E values” are given.
Example 1:
Rx Dexamethasone sodium phosphate 0.1%w/v
Purified water q.s. 30ml
Prepare an isotonic solution
Step-I: Reference solution: 0.9%w/v NaCl solution
100mL NaCl solution contains 0.9g NaCl
\ 30mL NaCl solution contains NaCl = 0.27g NaCl
The sodium chloride equivalent of the drug, E = 0.18
It means 1g drug is equivalent to 0.18g NaCl
Step-II: Contribution of the drug
30mL 0.1%w/v solution contains 0.1%w/v x 30mL dexameth. sod. phosph.
= = 0.03g drug
1g drug is equivalent to 0.18g NaCl
\ 0.03g drug is equivalent to 0.18 x 0.03g NaCl = 0.0054g NaCl
Step-III: Contribution of reference solution – Contribution of actual solution
= 0.27g NaCl – 0.0054g NaCl = 0.2646 g NaCl » 0.265g NaCl
N.B. Let us summarize the whole calculation
Ingredient
Quantity in 30mL
Equivalent to sodium chloride
Dexamethasone sodium phosphate
0.1%w/v
0.03g
0.0054g
Sodium chloride
0.2650g
Water q.s.
30mL
Total
0.2700g
Ans: To a 30ml solution of 0.1%w/v dexamethasone sodium phosphate, sodium chloride required is 0.265g to produce an isotonic solution.
DISPLACEMENT VALUE
Definition: The amount of drug(g) that displaces 1 gram of the base is called the displacement value of the drug.
The displacement value is constant for a drug and a base.
Mathematical expression:
Displacement value of a drug
Example-I: Calculate the displacement value of Zinc oxide in cocoa butter suppositories containing 40% zinc oxide and is prepared in 1g mould. The weight of 8 zinc oxide suppository is 11.74g.
Solution:
Weight of 1 suppository of pure cocoa butter base, w1 = 1g
Weight of 1 suppository of 40%ZnO suppository = 11.74g/8 = 1.4675g.
Amount of ZnO present in 1 suppository = 40% of 1.4675g = = 0.40 x 1.4675g = 0.587g
Amount of cocoa butter present in 1 suppository, w2 = 60% of 1.4675g = = 0.60 x 1.4675g = 0.8805g
Therefore, amount of cocoa butter displaced
= Weight of 1 pure cocoa butter suppository – Weight of cocoa butter present in 1 zinc oxide suppository
= 1g – 0.8805g
= 0.1195g
From definition, the displacement value of zinc oxide
= 4.9 » 5 Ans.
