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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5320_Библиотеки_им_академика_М_И_Перельмана
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v
cv
×
×
mL
c wcw
22
×=×
https://t.me/med1917
Pharmaceutical Dosage Forms and Drug Delivery
This formula can be used to calculate the volume of solvent required to make a diluted solution. For
example, to dilute a 50% w/ v stock solution to make 200 mL of a 5% w/ v solution, c1 = 50, c 2 = 5, and
v2 = 200.
1
22
5 200
=
c
1
50
=
20=
20 = 180 mL to make a total of 200 mL of the diluted solution.
The measurements can also be carried out in weight rather than in volume for the stock and the diluted
solutions. Thus,
11
(5.4)
5.4.5 Mixing Solutions of Different Concentrations
It is commonly necessary to mix two products that contain the same solute but have different
concentrations. A convenient approach to solve these problems is the alligation method. Two kinds of
alligation methods are commonly used: alligation medial and alligation alternate.
5.4.5.1 Alligation Medial
5.4.5.1.1 For Two Ingredients
is multiplied by its amount (e.g., quantity in grams) to obtain the product of each ingredient. The products
of all ingredients and their quantities in the original formula are added together separately. Dividing the
sum of products by the sum of quantities in the original formula gives a quotient, which represents the
mixed with 24 g of a 40% w/ v sucrose solution, one would write the alligation medial method as indicated
in Table 5.1
5.4.5.1.2 For More Than Two Ingredients
This method is also applicable to more than two ingredients. For example, to calculate the strength of the
TABLE 5.1
Alligation Medial Method for Two Ingredients
Ingredient Strength (% w/ w) Quantity (g)
A 10 12 120
B 40 24
Sum 1080
Product of Strength and
Quantity (% w/ w * g)
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Pharmacy Math and Statistics
TABLE 5.2
77
Alligation Medial Method for More Than Two Ingredients
Product of strength and
Ingredient Strength (% w/ w) Quantity (g)
A 10 12 120
B 40 24
C 5 180
Sum 72
TABLE 5.3
quantity (% w/ w * g)
Alligation Alternate Method for Two Ingredients
w/ v sucrose solution, one would write the alligation medial method as indicated in Table 5.2. Working
5.4.5.2 Alligation Alternate
5.4.5.2.1 For Two Ingredients
This method can be used to calculate the amount of a diluent, solute, or different concentration product
that would need to be added to a given concentration product to make a new concentration preparation.
The number of parts required for the lower- and higher- concentration preparations to make the targetconcentration preparation is obtained by constructing a matrix and doing the calculation, as shown in
Table 5.3.
Thus, subtracting the target concentration from the lower concentration gives the target amount of the
higher- concentration preparation, and subtracting the higher concentration from the target concentration gives the target amount of the lower- concentration preparation. Thus, the total amount of the target
concentration preparation that would be prepared can be obtained by adding together the target amounts
of higher- and lower- concentration preparations needed. Suppose the required amount of the target concentration preparation differs from the amount obtained by the formula. In that case, the principles of
proportion discussed earlier can be used to calculate the quantities needed for the required total amount
of the target- concentration preparation.
35
emL==
x
35
emL==
x
Conv
==
714
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78
TABLE 5.4
Pharmaceutical Dosage Forms and Drug Delivery
An Example of Alligation Alternate Method for Two Ingredients
For example, to prepare 200 mL of a 12% w/ v sucrose solution using a 40% w/ v and another 5% w/ v
sucrose solution, one would write the alligation matrix as shown in Table 5.4.
Thus, combining 7 mL of 40% w/ v solution with 28 mL of 5% w/ v solution would give 7 + 28 = 35 mL
of 12% w/ v solution. To make 200 mL of 12% w/ v solution, one would use the principles of proportion
as follows:
For the quantity of 40% w/ v solution,
7
mL
mL
200
mL
mL
Henc
7
×=
x,
35
200 40
For the quantity of 5% w/ v solution,
28
.
mL
mL
200
mL
mL
Henc
28
×=
x,
35
200 160
Alternatively, a conversion factor could be derived for the calculation:
200
35
mL
mL
. .
5
.
ersionfactor
The required quantities of low- and high- concentration solutions can then simply be obtained by multiplying their quantities obtained by the alligation formula by this factor. Thus, the quantity of 40% w/
v solution required = 7 × 5.714 = 39.998 = 40 mL. Therefore, the quantity of the 5% w/ v solution
5.4.5.2.2 For More Than Two Ingredients
The alligation alternate method can be used for more than two ingredients by pairing off the values of
one higher (than the desired) strength ingredient with two lower (than the desired) strength ingredients,
or vice versa. This is illustrated by the following example:
To prepare a 17.5% w/ w solution using a 10% w/ v, a 40% w/ v, and a 5% w/ v sucrose solution, one
would write the alligation alternate method, as shown in Table 5.5.
Thus, combining 20 mL of 40% w/ v solution with 22.5 mL of 10% w/ v solution and 22.5 mL of a 5%
The alligation alternate method for more than two ingredients can use any pairing of higher (than the
desired) strength ingredient(s) with lower (than the desired) strength ingredient(s). The pairings can be
any number, depending on the number of ingredients.
MW
acceptableerror
×
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Pharmacy Math and Statistics
TABLE 5.5
79
An Example of Alligation Alternate Method for More Than Two Ingredients
The alligation methods are applicable to all forms of preparations, including powders. In addition, the
alligation method can also be used for calculating the required quantities for dilution of a preparation with
the solvent or diluent alone by making the concentration of the lower- concentration preparation zero.
5.4.6 Aliquot Method of Dilution
There might be an instance when the drug quantity needed for dispensing is too small to be measured
using the available instruments. A pharmacist can achieve precision in measurement beyond the
instrument’s capacity by calculating and measuring in terms of aliquot parts. This occurs frequently
when compounding is required for extremely potent pharmaceuticals intended for neonates. The aliquot technique is applicable to both solids and liquids. Certain terminologies associated with the aliquot
method are:
Aliquot: Aliquot that are fractions, portions, or parts contained exactly a number of times in another.
Minimum measurable quantity (MMQ)
measured using a piece of certain equipment.
Acceptable error: Acceptable error is the permissible deviation in the error rate of a measuring
equipment based on its accuracy and established standards for the purpose. A 5% error margin is
often deemed acceptable in pharmaceutical calculations.
5.4.6.1 Dispensing Solid Using Aliquot Method
In this technique, an excess amount of the drug substance is weighed to ensure the balance’s sensitivity is
ture is weighed to contain the required quantity of the substance.
Minimum weighable quantity (MWQ):
The minimum amount of powder that can be precisely weighed using a certain prescription balance.
It can be calculated as:
sensitivityrequirement %
Q
=
100
Example
1.
MW
×100%
MW
×
5
%
MW
×100%
MW
mg
×
4
%
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80
Pharmaceutical Dosage Forms and Drug Delivery
Solution
Using the equation 5.27.
sensitivityrequirement
Q
=
6 100
Q
acceptableerror
mg
%
%
mg=
=
120
This indicates that a quantity less than 120 mg cannot be weighed on this balance without
surpassing the acceptable error percentage.
2. A pharmacy has a prescription balance, with a sensitivity requirement of 3 mg. If the pharmacy
prescription balance?
Solution
Using the equation 5.27.
sensitivityrequirement
Q
=
acceptableerror
%
mg
3 100
Q
=
%
=
75
Therefore, the minimum quantity you can weigh on this prescription balance is 75 mg without
exceeding the acceptable error of 4%.
Aliquot method
1.
percentage.
2.
ingredient.
3. Determine the total quantity of the active ingredient required for the aliquot by multiplying the
desired quantity of the active ingredient by the same multiplication factor.
4. To calculate the total quantity of the mixture, multiply the weight of the excipient or diluent by
the same multiplication factor.
5. Calculate the required amount of diluent by subtracting the calculated quantity of active
components from the overall mixture.
Mix the calculated active drug and diluent and mix well.
7. By applying the ratio- proportion approach weigh the mixture to match the initially requested amount.
Example problem
1.
balance in the pharmacy is 120 mg. Lactose as a diluent is available in the pharmacy. Calculate
mixture will have the prescribed amount of drug in it?
120
5
24
mg
52
×=
To
actor=×
To
=×=
Diluentmgmgmg=−=2 880 120 2 760,,
2
,,
gm+=
=
24
mg
()
+
()
()
aliquot
()()
()
()
120
()
,mg
2
880
−= +=
()
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Pharmacy Math and Statistics
81
Solution
It can be solved by two methods:
Method I
Step 2. Calculate the multiplication factor
mg
Multiplication factor =
=
Step 3. Determine the total quantity of the active ingredient required.
Amount of drug needed =
4 120mg mg
Step 4. Calculate the total quantity of the mixture.
talmixture MWQmultiplicationf
talmixture mg mg
120 24 2 880,
Step 5. Calculate the required amount of diluent.
760 120 2 880
mg diluentmgdru
Step 7. Apply the ratio- proportion approach.
2 880
,
=
120
Each 120 mg aliquot of the mixture contains 5 mg of the drug.
Method II. Apply the following equation:
A Weight of drugin drug diluentmixture
B Weight of drug dilu
880 120 2 760 120 2
,, , mg mg mg diluentmgdrug
+ eentmixture x
5
mg mg
==
x
120
mh
C Weight of drug in
=
D Weightof aliquo
2 880
tt
Weigh 120 mg aliquot from the mixture that contains 5 mg Atenolol
MMQ
×
MMQ
%
mL
×
5
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82
5.4.6.2 Dispensing Liquid Using Aliquot Method
Pharmaceutical Dosage Forms and Drug Delivery
Similarly, we can apply the aliquot method to measure the liquid when the quantity required is less than
Example problem
1. If a 1 mL measuring cylinder has a volume mark every 0.01 mL and your acceptable error rate is
sensitivityrequirement %
=
acceptableerror%
001 100
=
%
100
=
02..
This means that this measuring cylinder can accurately measure a minimum volume of 0.2 mL
with an error margin of less than 5%.
2. A prescription requires 0.2 mL of olive oil. The smallest measuring cylinder in the pharmacy is
a 10- mL graduate calibrated in units of 0.5 mL. How can you acquire the necessary quantity of
olive oil using the aliquot approach with ethanol as the diluent?
Solution
Step 1. Choose a multiple of the desired quantity that meets the necessary precision requirements
for measurement. In this case, 0.5 mL is the minimum quantity that can be measured using this
cylinder.
If 5 is chosen as the multiplication factor, then:
Measure 5 × 0.2 mL, or 1 mL of olive oil
Step 2. Calculate the quantity of diluent (ethanol) required by using the multiplication factor:
Diluent + drug mixture required = 1 mL × 5 = 5 mL
Step 3. Subtract the quantity of olive oil from the diluent (ethanol).
Step 4. Mix the quantity of olive oil and the diluent.
Step 5. Aliquot 1/ 5 of the dilution, or 1 mL, which contains 0.2 mL of olive oil.
3. Using a 25 mL graduated cylinder calibrated in 5 mL units, describe how 1.25 mL of a drug solution could be measured using the aliquot method. Use water as a diluent.
Solution
Find a multiplication factor that can be measured using the graduated cylinder.
Multiplication factor = 1.25 mL × 4 = 5 mL
Amount of water + drug solution required = 5 mL × 4 = 20 mL
Mix the 5 mL drug solution and 15 mL water
Aliquot 5 mL, which contains 1.25 mL of drug
21
21
21
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Pharmacy Math and Statistics
5.4.7 Tonicity, Osmolarity, and Preparation of Isotonic Solutions
83
from the solution with the lower solute concentration to the solution with the higher solute concentration.
membrane. This phenomenon is called osmosis. The pressure of the solvent involved in this phenomenon
is termed osmotic pressure. A solution containing a nonpermeable solute creates pressure for the inward
membrane.
Tonicity is the osmotic pressure of two solutions separated by a semipermeable membrane. Tonicities
hypotonic, while solutions that exert higher
hypertonic. Hypotonic solutions have lower and hyper-
the same osmotic pressure are termed isosmotic, while a solution with the same osmotic pressure as a
isotonic.
concentration of a solute without referring to another solution. An osmole is the amount of a substance
that represents the number of moles of particles that it forms in a solution. For a nondissociating sub-
(molecular weight) of dextrose.
Similar to the concept of molarity, osmolarity
of glucose dissolved in 1 L of solution. Similar to the concept of molality, osmolality
such as milli and micro. Thus, a commonly used term is milliosmole (abbreviation: mOsmol), which
represents 1/ 1,000th of an Osmol. Moreover, while osmole represents the quantity of solute in grams,
Osmol represents the solute concentration in a solution.
osmoles and osmolarity of such a solute are calculated by multiplying with the number of particles formed
on dissociation and the fractional degree of dissociation of a substance in solution. Thus, assuming complete dissociation, NaCl, CaCl2, and FeCl3 form 2, 3, and 4 particles in solution. Thus, a 1 mM NaCl,
CaCl2, or FeCl3 represents their 2, 3, or 4 mOsmol solution, respectively. Assuming an 80% degree of
dissociation for dilute solutions, 2 M of NaCl, CaCl2, and FeCl3 solutions represent:
be measured in the laboratory using an osmometer.
Tonicity is an important concept in the administration of ophthalmic and parenteral solutions.
-
tonic solutions are relatively inconsequential since the volume of the administered solution is much
80
= .Osmolof NaClsolution
100
36×+
80
×+ +
10080100
80
× +++
1008010080100
= .Osmolof CaCl solution
52
= .Osmolof FeCl solution
68
2
3
18
05
50
°
°
==×=
x
18
9
..
°×
°
==
×
×
x
Drug
100
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84
Pharmaceutical Dosage Forms and Drug Delivery
administration of hypertonic solutions tends to be more tissue- damaging and painful than the administration of hypotonic solutions. Nonetheless, isotonic solutions are better tolerated by patients than the
extreme of tonicity.
Preparation of isotonic solutions requires the use of one of the colligative properties of solutions.
Colligative properties are the solution properties that depend on the number of molecules of solvent in
a given volume of solution but are independent of the properties of the solute. These properties include
lowering vapor pressure, elevation of boiling point, osmotic pressure, and depression of freezing point of
a solution with increasing solute concentration. Of these, the depression of freezing point is conveniently
used to calculate the amount of solute required to prepare an isotonic solution.
calculate the amount of glucose (molecular weight: 180 g/ mol) required to prepare an isotonic solution
make 1 L of isotonic glucose solution, the amount of glucose required (x) can be calculated as:
180
6
.
C
2
.
C
g
g
Therefore
x
,
180
052
.
186
.
This corresponds to 5% w/ v glucose solution. The commonly available dextrose solution for intravenous (IV) administration has this concentration. A similar concentration for an electrolyte, such as
sodium chloride, should take into consideration the dissociation constant of the solute and the number
of species produced in the solution. Thus, assuming NaCl in weak solutions is about 80% dissociated,
the total number of solutes in the solution would be 1.8 times the number of molecules added. This (1.8)
dissociation factor (abbreviation: i) is used to calculate isotonic concentrations of electrolytes. Thus, to
make a 1 L isotonic NaCl (molecular weight: 58 g/ m) solution, the amount of NaCl required (x) can be
calculated as:
618
..
052
.
C
58 58 052
g
C
g
Therefore
.
x
,
18618
=
This corresponds to 0.9% w/ v NaCl solution, which is commonly available as an isotonic solution for
experiments involving living cells and tissues. These calculations show that 50 g/ L of glucose solution is
isotonic to 9 g/ L of NaCl solution. Therefore, 50 g of glucose is tonic equivalent to 9 g of NaCl in quan-
tities of solutes. The tonic equivalence of two substances represents the amounts that would produce the
same osmotic pressure. Thus, the quantity of any substance divided by its dissociation factor, i, represents
its tonic equivalent quantity to any other substance. This principle is used in the preparation of isotonic
solutions by the addition of NaCl to hypotonic drug solutions to increase the tonicity to the physiological
equivalent of 0.9% w/ v NaCl. Using the above conversion of tonic equivalents, NaCl equivalents (E
values) of various substances are known in the literature. The number of grams of all ingredients in a prescription is multiplied by their E values and added to determine the osmotic equivalent of NaCl amount
represented by the substances. In addition, the amount of NaCl required to make a 0.9% w/ v solution
of the same volume as the prescription is determined. Subtracting the former from the latter gives the
amount of NaCl needed to make the solution isotonic. Any substance other than NaCl, such as dextrose,
can also be used to increase the tonicity of a solution by dividing the amount of NaCl needed by the NaCl
equivalent of the other substance.
determine the amount of drug in 10 mL of solution.
amount g=×=
3
10 03.
Ev
==
216
ng=× =
To
100
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Pharmacy Math and Statistics
85
The NaCl equivalent (E value) of pilocarpine nitrate (molecular weight 271, dissociates into 2 ions,
and i value = 1.8) can be read from the literature or calculated as:
58 518
alue
./.
271 18
/.
0
.
Now, we multiply the E value with the drug amount in the solution to get NaCl equivalents represented
by the drug amount in the solution:
NaClequivalentin prescriptio
03 0 216 0 0648.. .
This is the amount of particles in solution equivalent to NaCl, which must be subtracted from the
amount of NaCl needed to make an isotonic solution of the same volume as the prescription (i.e., 10 mL).
This is calculated as:
talamountof NaClneededforisotonicityg=×=
09
10 009..
If a prescription contains multiple components, the NaCl equivalent for each component is calculated
separately and added together to make the total NaCl equivalents in the prescription. This total amount is
then subtracted from the total NaCl that would be needed for the isotonicity of the volume of prescription
to obtain the amount of NaCl that must be added to the prescription.
5.5 Clinical Dose Calculations
The dose of a drug represents the amount of the drug substance that a patient must take at one time. This
amount is designed with the expectation of producing the optimum therapeutic effect while minimizing
the unwanted side effects. In the current pharmacokinetic paradigm, the designed therapeutic dose for a
patient is usually based on the desired target concentration of the drug substance in the patient’s central
changes, for example, can include patient- to- patient differences in body weight, body surface area (BSA),
drug required for an average 180- lb adult with normal body functions. The drug’s dose for an individual
to the drug substance.
5.5.1 Dosage Adjustment Based on Body Weight or Surface Area
In many cases, the target dose is expressed in terms of BSA or body weight. For example, meperidine
hydrochloride (Demerol®
contrast, while isoniazid has a recommended daily dose of 450 mg/ m2 BSA/ day to be administered in a
single dose. Therefore, the daily dose is calculated based on the patient’s weight or BSA and divided by
the number of doses per day to determine an individual dose amount. A set of doses administered over a
period of time as a part of a treatment plan is termed dosage regimen.
For example, for a patient of 180 lb body weight, the daily dose of meperidine hydrochloride would be
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