Pharmaceutical Calculations and Metrology
Syllabus:
Metric and imperial systems of weights and measures used
in prescriptions, posology, calculations of doses for infants, children, adults
and elderly patients, reducing and enlarging formulae; percentage solutions,
alligation methods; proof spirits; calculations involving alcohol dilutions; pH
and buffer solutions, isotonic solutions; displacement value and calculations
involving radio-isotopes.
Reference Books:
1.
Dispensing Pharmacy, R.M.Mehta
2.
Cooper & Guns Dispensing
3.
Pharmaceutics-II, A.K. Gupta
4.
Remington’s Pharmaceutical Sciences.
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 :
1
BUFFER CAPACITY
BUFFER 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
|
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.
CALCULATIONS INVOLVING RADIOISOTOPES
Disintegration
Rate of a Radioisotope
Radioisotope
nucleus ®
Daughter nucleus
The rate at which the
radio-isotopes are degrading =
|
where, N is the number of
radioactive nuclei left to be disintegrated at time t
|
It is found that the degradation
follows first order kinetics, i.e. ,
|
where l = decay constant of the
radionucleide
|
So, or,
Integrating both sides we get
or, or, lnN = lnN0 – l t or,
or,
N/N0 = or, N =
N0
N0 = initial number of
radioactive nuclei
N = number of radioactive nuclei at
time t
Half
life of a radioactive nuclei
The time taken for half of the
radioactive nuclei to disintegrate (i.e. for the activity to fall to half of
its original value) is known as the half-life (t1/2)
At time, t = 0 N = N0.
At time, t = t1/2 N = N0/2
Therefore, , or,
or,
ln (1/2) = –l t1/2. or,
–ln 2 = –l
t1/2. or,
t1/2 =
Units
of radioactivity
One g of radium was selected as the
unit of radioactivity and was called Curie.
From 1 curie (Ci) radium 3.7 x 1010 numbers of nuclei
disintegrates per second.
So 1
Ci = 3.7 x 1010 dnps dnps
= disintegrating nucleus per second
Problem:
The
activity of a sample solution of 131I was 500 mCi (microcurie) /ml at noon on Monday. Calculate its
activity at 4.0 p.m. on Thursday. (Half life of 131I = 8days)
Solution:
t1/2 = 8 days = 8
x 24 hours = 192 hours
From equation, t1/2 = , we get l = = = 0.00361 hr –1.
From equation lnN = lnN0
– l t
In this problem N0 = 500 mCi /ml
t
= the time from 12 noon Monday to 4 p.m. Thursday = 3x24 + 4 = 76 hrs
\ lnN = ln 500
– 0.00361 x 76
=
6.218 –
0.2743
=
5.994
or,
N = e5.994 = 381..1
Ans.
At 4 p.m. Thursday the radioactivity will be 381.1 mCi / ml.
Questions for Test-II
& Final Examination.
1. Describe
the factors affecting the dose of adrug. [12]
2. Calculate
the quantity of sodium chloride required for 500ml of 0.9% solution.
3. Prepare
600ml of 60%v/v alcohol from 95% v/v alcohol. By alligation method.
4. Find
the strength of 95%v/v alcohol in terms
of proof spirit.
5. What
is proof spirit. Calculate the real strength of 300op and 400up.
6. 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.
7. Define
an isotonic, hypertonic and hypotonic solution.
8. Define
displacement value. 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.
9. Write
an equation for disintegration of radioisotopes. Determine the half life of a
radioactive nuclei from this equation. What is the unit of radioactivity?
10. The
activity of a sample solution of 131I was 500 mCi
(microcurie) /ml at noon on Monday. Calculate its activity at 4.0 p.m. on
Thursday. (Half life of 131I = 8days)