The concentration of the components in a pharmaceutical preparation is based on its quantity relative to the total quantity of the preparation .
The aim : to learn pharmacy students how to prepare pharmaceutical preparation of wanted concentration from another pharmaceutical preparation of different concentration.
Contents :
1.
Dilution & concentration of liquids:
2.
stock solution
3.
Dilution of alcohol
4.
Dilution of acids
5.
Dilution of solids & semisolids
6.
Trituration
7.
Alligation
Dilution:

By addition of diluents .

By admixture with solutions or mixture of lower strength .
Concentration :
 By addition of active ingredient
 By admixture with solutions or mixture of greater strength .

Evaporation of the diluents.
Calculating the percentage or ratio strength of the solution made by diluting or concentration a solution of given quantity & strength by equation:
1 ..
.
Inverse proportion .
2 ..
.
The equation: ( Q1  C1 = Q2  C2)
(1 st quantity  1 st concentration) = (2 nd quantity  2 nd concentration).
1

3 ..
.
By determining the quantity of active constituents ( solute ) needed & then calculating the quantity of the available solution
(usually concentrated or stock solution ) that will provide the needed amount of constituents.
Dilution & concentration of liquids:
Example : If 500 ml of 15% (v/v) solution of methyl salicylate in alcohol are diluted to 1500 ml . What will be the percentage strength ?
1500 ml
500 ml

15 % x %
x= 5%
Or :
Q1  C1 = Q2  C2
500 ml  15%=1500 ml  x % x= 5%
Or :
500 ml of 15% (v/v) solution contains 75 ml of solute
1500 ml x
75 ml

5 %

100 % x %
Ex.: If 50ml of 1:20 w/v solution are diluted to 1000 ml , what is the ratio strength (w/v)?
1:20 = 5%
1000 ml
 x
50 ml

0 .
25 %
5 % x

%
1 : 400
Or:
1000 ml
 x
50 ml

1 / 400
1
20
 x
1 :

400
Or:
Q1  C1 = Q2  C2
2

50ml  5% = 1000ml  x% x =0.25% = 1:400
Or:
50ml of a 1:20 solution contains 2.5g of solute
2 .
5 ml
1 g

1000 ml xml x

400 ml

1 : 400
Ex: How many grams of 10 (w/w) ammonia solution can be made from
1800g of 28% w/w strong ammonia solution?
10 %

1800
28 % x

5040 g x g
Or : Q1  C1 = Q2  C2
1800  28% = x g  10% x = 5040g
Ex: How many milliliters of a 1:5000 w/v solution of drug can be made from 125ml of a 0.2% solution ?
1:5000 = 0.02%
0 .
02 %

125 x
0 .
2 %

1250 ml ml xml
Or :
0.2% = 1:500
1
5000 
125 x
1
500

1250 ml ml xml
Or:
1 g
0 .
25 g

5000 ml xml x

1250 ml
3

Stock solution are concentrated solution of active ( e.g. drug ) or inactive ( e.g. colorant ) substances & are used by pharmacists as a convenience to prepare solution of lesser concentration (strong solutions from which weaker once may be made conveniently ) .
Examples :
How many milliliters of a 1:400 w/v stock solution should be used to make 4L of a 1:2000 w/v solution ?
4L=4000ml
1:400=0.25%
1:2000=0.05%
0 .
25 %

0 .
05 % x

800 ml
4000 ml xml
Or x
1
1 /
/ 400


2000
800 ml
4000 ml xml
Or 4000 ml of 1:2000 w/v solution requires 2g of active constituent ;
1 g
2 g

400 ml xml x

800 ml
Or: Q1  C1 = Q2  C2
4000ml x 0.25%= x x 0.05% x =800 ml
Ex: How milliliters of a 1:50 stock solution of phenylephrine HCl should be used in compounding the following prescription ?
Rx: Phenylephrine HCl 0.25%
Rose water ad 30 ml
Sig. for the nose.
1:50 = 2 %
4

2 %

30
0 .
x
25 %

3 .
75 ml ml xml
Or :30g x 0.0025=0.075 g of Phenylephrine HCl needed
1:50 means 1g in 50ml of stock solution
1 g

50 ml
0 .
075 x

3 .
xml
75 ml
Calculation of active constituent:
Ex: 500 ml of 15% v/v solution in alcohol are diluted to 1500ml . What will be the % v/v?
Q1 x %1 = Q2 x % 2
500 x 15 % = 1500 x % after dilution x

500 x 15
1500 x

5 % v / v or 1500 ml 15%
500 ml x %
x =5 % v/v
Ex: How many grams of 10% w/w Ammonium can be made from 1800 g
, 28% w/w Ammonium ?
1800 g x 28 % = x g x 10 %
X = 5040 g
Ex: How many mls of water should be added to 375 ml of solution containing 0.5g of benzylkonium Cl to make 1:5000 solution ?
5000 x 0.5= 1 x x x =2500ml  2500 -375 = 2125 ml of water added
5

Ex: How many mls of water should be added to 450 ml of solution 1:900 w/v of methyl salicylate to make solution of 1:3000 w/v ?
1:900 = 0.11% , 1:3000 =0.033%
450ml x 0.11 = V2 x 0.033
V2 1500 ml
1500-450= 1050 ml of water added
Dilution of alcohol:
When water & alcohol are mixed , there is a physical contraction such that the resultant volume is less than the total of the individual volumes of the two liquids .
Ex: how much water should be mixed with 5000 ml of 85% v/v alcohol to make 50% v/v alcohol ?
50 %

5000
85 % x

8500 xml ml ml
Therefore, use 5000ml of 85% v/v alcohol & enough water to make 8500 ml .( 8500-5000=3500 ml water added).
Ex: How many mls of 95% v/v alcohol & how much water should be used in compounding the following prescription :
Boric acid 1g
Alcohol 70% 30ml
Sig. Ear drops. ?
70% x 30ml = 95% x Xml
X= 22.1 ml of alcohol % dissolve 1g of boric acid then complete the volume to 30 ml by water .(30ml-22.1ml=7.9 ml of water )
Dilution of acids :
The strength of an official undiluted (concentrated) acid is expressed as percentage w/w .
6

Ex: How many mls of 85% w/w phosphoric acid having a specific gravity of 1.71 should be used in preparing 1 gallon of ¼% w/v phosphoric acid solution to be used for bladder irrigation ?
1 gallon=3785 ml
3785 x 0.0025= 9.46 g of phosphoric acid (100%) in 1 gallon
85 %
100 %

9 .
46 g xg x

11 .
13 gof 85 % phosphoric acid
11.13g of water measures 11.13 ml
11.13 ml /1.71=6.5 ml
Dilution & concentration of solids & semisolids :
Dilution of solids in pharmacy occurs when there is need to achieve a lower concentration of an active component in a more concentrated preparation .
Ex: how many grams of opium containing 15% w/w of morphine & how many grams of lactose should be used to prepare 150g of opium containing 10% w/w of morphine?
15 %
10 %

150 g xg x

100 gof 15 % opium
150g -100 g= 50g of lactose.
Ex: if 7.2%w/w active ingredient contained in crude plant & 21.6 % water , what is the percent of active ingredient after the drying ?
100 g – 21.6 = 78.4 g after drying
78 .
4 g
 x
100

9 .
2 %
7 .
2 % x %
Ex: how many grams of zinc oxide should be added to 3200g of 5% zinc oxide ointment to prepare an ointment containing 20% of zinc oxide ?
3200g x 0.05 =160g of zinc oxide of 5%
7

3200g – 160g =3040g of base in 3200g of 5% ointment
80 %
20 %

3040 g xg x

760 gof zinc oxide in 20% ointment
Because the 5% oint. Already contains 160g of zinc oxide
760 – 160 =600g
8