Pharmacological calculations
Solutions
• Liquid preparations containing solid substances dissolved in a
fluid medium are called solution. Preparation of a solution may
be facilitated by shaking, vortexting or by sonication.
Types:
• Stock solution and its dilution
• Percentage solution
• Moles/Milli moles
• Equivalent/Milli equivalent
• Osmolar/Milli osmoles
Stock Solutions
• Stock solutions are solutions of known concentration that are
prepared by the pharmacist or technician for convenience in
dispensing. They are usually strong solutions from which
weaker ones may be made conveniently. When correctly
prepared, these solutions enable the pharmacist to obtain
small quantities of medicinal substances that are to be
dispensed in solution.
Formula
(Quantity1) ×(concentration1) = (Quantity2) ×(concentration2)
C1 × V1= C2 × V2
C1= concentation of stock solution
V1= volume of stock solution
C2= concentration of required solution
V2= volume of required solution
• These two rules, wherever they may be applied, greatly
simplify the calculation:
1. When ratio strengths are given, convert them to percentage
strengths before setting up a proportion. It is much easier to
solve using a decimal or percent, like 10 (%):0.2 (%) than a
fraction represented by the ratio like 1⁄10 :1⁄500.
2. Whenever proportional parts enter into a calculation, reduce
them to lowest terms. Instead of calculating with a ratio like 25
(parts):75 (parts),simplify it to 1 (part):3 (parts).
Example 1
• If 500 mL of a 15% v/v solution of methyl salicylate in alcohol is
diluted to 1500 mL, what is the percentage strength v/v?
C 1 × V1= C2 × V2
500ml × 15 % =1500ml × %
X= 5%
Example 2
• How much 10% w/w (in grams) ammonia solution can be made
from 1800 g of 28% w/w strong ammonia solution?
C1 × V1= C2 × V2
1800g × 28%= Xg × 10%
X= 5040 g
Example 3
How much (in milliliters) of a 1% stock solution of a certified red dye
should be used in preparing 4000 mL of a mouthwash that is to
contain 1:20,000 w/v of the certified red dye as a coloring agent?
1:20,000 = 0.005 %
C1 × V1= C2 × V2
X × 1 % = 4000ml × 0.005(%)
X=20ml
Example 4
• Hydrocortisone is available in a vial of 100 mg in 2 mL. You are to
prepare 24 mL of a 5-mg/mL hydrocortisone dilution using the
available stock vials. How much hydrocortisone (in milliliters) and
how much diluent (in milliliters) will you need?
C1 × V1= C2 × V2
5 mg/mL =0.005 g/mL, a 0.5% solution
• 100 mg/2 mL =0.1 g/2 mL, a 5% solution
• X × 5% =24ml × 0.5%
• X=2.4ml hydrocortisone
• 24ml-2.4ml=21.6ml diluent
Percentage Solutions
• One part of any substance mixed in 100 parts of another
substance makes one percent solution.
• Percent has no unit.
• It is expressed as %.
• It may be weight by volume (W/V), weight by weight (W/W) or
volume by volume (V/V)
Example:1
Calculate the amount of NaCl required to prepare 1.5
L(1500ml) of .015% solution.
.015% of sodium chloride solution means 0.015 g of NaCl
dissolved in 100ml
If 100ml of soln contain NaCl = 0.015gm
1ml of soln contain NaCl = 0.015/100 gm
1500 ml of soln will contain NaCl = 0.015/100 × 1500
=0.225 gm or 225mg
Weigh 225 mg of NaCl and dissolve in distilled water to make
upto 1.5 L solution.
Example :2
Calculate the amount of pilocarpine required to inject in a 1.25Kg
rabbit when the dose of drug is 10mg/kg body weight and dosage
form is 1% W/V.
Data: Weight of rabbit =1.25 kg, Dosage form=1% W/V
Dosage of drug =10mg/kg
Solution: Drug to be injected =dose per unit × weight of a rabbit
=10 ×1.25=125mg=0.0125 g
As dosage form is 1% means 1g of drug dissolved in 100ml
=100 ×0.0125
Result: Drug to be injected is 1.25ml
Example :3
Calculate the amount of NaCl required to prepare 500ml o
frog’s Ringer’s saline solution. The composition of solution is
0.65%
Composition of solution = 0.65%
It means 0.65gm dissolved in 100ml
100ml of solution contain=0.65g
1ml of solution contain =0.65/100
500ml of solution contain =0.65/100 ×500=3.25 g of NaCl
Fractional solutions
These are the solutions in which relative quantities of solute
and solvents expressed as ratio e.g 1:100, 1:1000 etc. It means
1gm of a drug is dissolved in 100 or 1000ml of distilled water
respectively.
Example :1
Calculate the concentration of acetylcholine in 50ml organ bath
if we dissolve 1.5ml of 1:6250 stock solution.
1:6250 means 1g of acetylcholine dissolved in 6250 ml of distilled water.
If 6250ml of solution contain Ach=1g=1000mg
1ml of solution will contain Ach =1000/6250mg
1.5 ml of solution will contain Ach= 1000/6250 × 1.5=0.24 mg
=240μg
If 50 ml organ bath have Ach = 0.24mg
1ml will contain =0.24/50mg
= 0.0048mg
=4.8μg/ml
Example :2
Calculate the volume of 2% atropine required to prepare 50ml of 1:750 solution.
1:750 means 1gm of atropine dissolved in 750ml of solvent
If 750ml of solution contain atropine =1gm
1ml of solution contain atropine =1/750 × 50=0.0679 g
1gm of atropine dissolved in 100 ml of soln.=100/2
=100/2×0.067
= 3.32 ml
We will use 3.32 ml of 2% atropine and dilute it upto 250ml to obtain the strength
1:750.
Molecular or Molar solutions
• A mole is defined as the molecular weight of a drug expressed in
grams.
• One mole of a solution means one gram molecular weight of a drug
dissolved in one liter.
• Gram molecular weight of any substance contain 6.02× 1023
particles
For example 180g of glucose molecule,23 g of sodium ion,35.5 g of
chloride ion all are equal to 1 mole or 6.02× 1023 molecules or
particles. It is called Avogardro’s number.
1millimole =1/1000M or 1Millimole is =0.001M(1 formula weight in
milligram)
1 Micromole = 0.001 mM ( 1 formula weight in microgram)
Example :1
• Calculate the amount of CaCl2 required to make 150ml of
8.5mMol solution (Molecular weight of CaCl2 =111)
If 1M of solution require CaCl2 =111g
1mM of solution require CaCl2 =111mg
8.5mM of solution require CaCl2 =111×8.5mg
=943.5mg
1000ml of 8.5 mM solution contain CaCl2 =943.5 mg
1ml of solution contains CaCl2 =943.5/1000mg
150ml of solution contains CaCl2 =943.5/1000 × 150mg
=141.525mg
Example :2
What is the millimole concentration of 2.5% solution of
KCl(MW=74.5)?
2.5% means 25gm in 100mL= 0.1 liters
Moles=wt in grams/Mol.wt
=2.5/74.5=0.0335×1000=33.35
mmol conc. = 33.35/0.1=333=333.5m mol
Equivalent (Eq)solutions
• An equivalent is the weight of a substance which combines with or
replaces one gram atomic weight of hydrogen ion.
• It can be calculated by dividing the molecular weight of a radical by
its valency.
• Gram equivalent weight of a solute dissolved in one litre of solvent
gives one Equivalent solution.
• One milliequivalent (mEq) is 1/1000 of an equivalent.
• For example Eq weight of K (which has a valency=1)
Eq wt =39/1 =39gm
mEq wt =39/1000=0.039 gm = 39 mg
Example :01
Calculate the amount of NaCl required to prepare 100ml solution
which contain 200mEq of Na (MW NaCl 58.5)
If 1Eq of NaCl solution contains =58.5 g
1mEq of NaCl solution contains =58.5mg
200 mEq of NaCl solution contains =58.5 × 200mg
=11700mg
If 1000ml of NaCl solution contain =11700mg
1ml of NaCl solution will contain =11700/1000mg
100 ml of NaCl solution will contain =11700/1000×100
=1170mg=1.17g
Osmolar solutions
• Osmole is gram molecular weight divided by number of particles or
ions into which a substance dissociates in solutions.
• Osmolarity is one of the colligative properties of a solution which
is based on the concentration of solute particles.
• One mole in case of non- electrolytes is equal to 1 Osmole.
• In case of electrolytes it is Mole × number of particles into which it
dissociates in solution.
• Milliosmole is 1/1000 of an Osmole.
• 1M of Glucose =1 Osm(glucose is nonelectrolyte)
• 1M of NaCl = 2Osm of NaCl (because one molecule gives two ions
in solution)
Example:01
• Calculate the amount of NaCl required to prepare 500ml solution having 300 milliOsmole
per liter concentration. What is the percentage of this solution?
Amount of soln required=500ml
Conc. of solution requires=300mOsm/L
Molecular weight =58.5
1 mole of NaCl = 2 Osm = 58.5 /2 g
1mmole = 2mOsm =58.5/2=29.25
1m Osmole of NaCl requires =29.25mg
300mOsm of NaCl will require=300×29.25=8775mg
1000 ml of NaCl require =8775mg
1 ml of NaCl require =8775/1000
500ml of NaCl require =8775/1000×500=4387.5mg=4.38g
1%=1g in 100ml
500ml contain NaCl =4.38gm
1 ml contain NaCl =4.38/500
100 ml contain NaCl =4.38/500 × 100=0.87%
Example:2
• How many grams of glucose are required to prepare 500ml solution
having 300 milliOsmole per litre concentration. What is the percentage of
this solution? (MW=180)
One molecule of glucose gives one ion when in solution form
1 mole of glucose=1 Osmole =180 g
1mM of glucose =1 mOsmole =180mg
If 1 mOsm of glucose require=180mg
300 mOsm of glucose will require =300×180mg=54000mg
1000ml of glucose soln require=54000mg
1ml of glucose soln require =54000/1000mg
500 ml of glucose solution require=54000/1000×500=2700mg=27g
27g of glucose,5.4%
