1. Dilution and Concentration
Chapter 10
Dilution and Concentration

2. Objectives Upon completion of this chapter, you will be able to:
Describe the relationship of active ingredients and diluents if the amount of active ingredient remains constant and the amount of diluent is increased or decreased
Determine the percent strength and ratio strength of a given product when the active ingredient remains constant and the amount of diluent is increased or decreased 2

3. Objectives (cont.)
Determine the volume of solution of a desired strength given a specified quantity of any given strength
Determine the volume of a specified stock solution needed to prepare a given solution
Determine the quantity of an active ingredient in a specified amount of solution needed to prepare a given solution
Define the alligation methods of problem solving 3

4. Objectives (cont.)
Utilize the alligation methods (alligation alternate and alligation medial) to determine the percent strength of alcohol mixtures
Utilize the alligation methods (alligation alternate and alligation medial) to determine relative amounts of components mixed together to make a mixture of a required strength 4

5. Definitions Diluent: Stock solutions:
A substance that is added to a pharmaceutical product to reduce the strength of the product. A diluent most often has no drug substance in it, sterile water and petrolatum for example.
Stock solutions:
Strong solutions from which weaker ones are made 5

6. Rules
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
2. Whenever proportional parts enter into a calculation, reduce them to lowest terms
them to percentage strengths before setting up a proportion
calculation, reduce them to lowest terms 6

7. Relationship Between Strength and Quantity
The amount of active ingredient remains constant; any change in the quantity of a solution or mixture of solids is inversely proportional to the percentage or ratio strength. (Or, as the volume increases the strength decreases.) 7

8. Methods of Problem Solving
1. Inverse proportion
2. The equation:
Q1 (C1) = Q2 (C2)
3. By determining the quantity of active constituent (solute) needed and then calculating the quantity of the available solution (usually concentrated or stock solution) that will provide the needed amount of constituent

9. Which one to use?
For most problems, the equation (#2) is the easiest to use
In some situations, when using vials or amps as your stock product, #3 would be the best method 9

10. Dilutions and Concentrations of Liquids
Example
If 500 mL of a 15% (v/v) solution of methyl salicylate in alcohol are diluted to 1500 mL, what will be the percentage strength (v/v)?
Q1 (C1) = Q2 (C2)
500 mL (15 %) = (1500 mL) (X %)
X = 5% (answer) 10

11. Example
If 50 mL of a 1:20 (w/v) solution of aluminum acetate are diluted to mL, what is the ratio strength (w/v)?
Use same formula
(50 mL) (5 %) = 1000 mL (X %)
X = 0.25% = 1:400 (answer) 11

12. Determine Amount of Solution of a Desired Strength
How many grams of 10% (w/w) ammonia solution can be made from g of 28% (w/w) strong ammonia solution?
Use the same formula
(1800 g)(28 %) = (X g) (10%)
X = 5040 g (answer) 12

13. Example
How many milliliters of a 1:5000 (w/v) solution of phenylmercuric acetate can be made from 125 mL of a 0.2% solution?
1:5000 = 0.02%
Use the same formula
(125 mL) (0.2 %) =(X mL) (0.02 %)
X = 1250 mL (answer) 13

14. Example
How many milliliters of a 1:400 (w/v) stock solution should be used to make 4 liters of a 1:2000 (w/v) solution?
4 liters = 4000 mL
1:400 = 0.25% :2000 = 0.05%
Use the same formula
X mL (0.25 %) = 4000 mL (0.05%)
X = 800 mL (answer) 14

15. Example
How many milliliters of a 1:400 (w/v) stock solution should be used in preparing 1 gallon of a 1:2000 (w/v) solution?
1 gallon = 3785 mL
1:400 = 0.25% :2000 = 0.05%
X (0.25%) = (3785 mL)(0.05%)
X = 757 mL (answer) 15

16. Using a Vial or Amp Stock
We are to prepare 30 mL of a 5 mg/mL oral phenobarbital solution using 1 mL vials with a concentration of 65 mg/mL. How much stock solution will be required? 16

17. Step 1: Calculate the amount of phenobarbital needed
30 ml x 5 mg/mL tells us we need 150 mg of phenobarbital for the solution

18. Step 2:
Calculate the volume of available stock needed. Our supply is 65 mg/mL, so
65 mg = 150 mg
1 mL X
X = 2.3 mL of phenobarbital will have to be drawn up

19. In order to use our formula we must calculate the percent strength of each ingredient 5 mg/mL = g/mL which is a 0.5% solution 65 mg/mL = g/mL = 6.5% solution X (6.5%) = (30 mL) (0.5%) X = 2.3 ml of phenobarbital required (answer)
In order to use our formula we must calculate the percent strength of each ingredient
5 mg/mL = g/mL which is a 0.5% solution
65 mg/mL = g/mL = 6.5% solution
X (6.5%) = (30 mL) (0.5%)
X = 2.3 ml of phenobarbital required (answer) 19

20. Determining Quantity of Active Ingredient in Specified Amount of Solution Given Strength of Diluted Portion

21. Example
How much silver nitrate should be used in preparing 50 mL of a solution such that 5 mL diluted to 500 mL will yield a 1:1000 solution?
1000 mL = 1 g
500 mL X g
X = 0.5 g of silver nitrate in 500 mL of diluted solution (1:1000), which is also the amount in 5 mL of the stronger (stock) solution, since the 50 mL and the 5 mL are the same strength 21

22. Example (cont.)
5 mL = g
50 mL X g
X = 5 g (answer) 22

23. Amount of Diluent Needed for Preparing Solution of Specified Lower Strength23

24. Example
How many milliliters of water should be added to 300 mL of a 1:750 (w/v) solution of benzalkonium chloride to make a 1:2500 (w/v) solution?
1:750 = 0.133% 1:2500 = 0.04%
Using our formula, recall that the Q2 always represents the FINAL volume
(300 mL) (0.133%) = X (0.04%)
X = or 1000 mL

25. Example (cont.) 1000 mL – 300 mL = 700 mL (answer)
This is the volume of our 0.04% solution; therefore, we must subtract the volume of the 0.133% solution (the amount we started with) from this final volume to determine how much water or diluent was added.
1000 mL – 300 mL = 700 mL (answer) 25

26. Alligation
Arithmetical method of solving problems that involves mixing of solutions or solids that have different percentage strengths 26

27. Alligation Medial Uses the weighted average
Used when the percent strength of each component is KNOWN
Express the percent strength by a decimal fraction
Multiply the decimal by the corresponding quantity
Add the products and divide by the total quantity of the mixture 27

28. Example
What is the percent strength of a product that is made up of 3000 mL of 40% alcohol, 1000 mL of 60% alcohol and 1000 mL of 70% alcohol?
40% = 0.4 x 3000 mL = 1200 mL
60% = 0.6 x 1000 mL = 600 mL
70% = 0.7 x 1000 mL = 700 mL
Add total volume = 5000 mL 28

29. Example (cont.) Add the products 1200 mL + 600 mL + 700 mL = 2500 mL
Divide 2500 mL by 5000 mL = 0.5 (change this to a percent) x 100 = 50%
Therefore by mixing these products the final product is 5000 mL of a 50% alcohol product 29

30. Alligation Alternate
This method calculates the number of PARTS of two or more components of a certain strength. The final proportion allows us to use proportional parts to make any amount of the product we wish.
Strength of the final product must be in-between the strengths of the components we are mixing 30

31. Example Mix a 95% product with a 50% product to get a 70% product
Subtract 70 from 95 and get 25
Subtract 50 from 70 and get 20
Now we have 20 parts of 95% and 25 parts of 50%
Remember to REDUCE 20:25 = 4:5

32. 32

33. Example (cont.)
So if we mix 4 parts of 95% and 5 parts of 50% we will get a 70% product
We can apply this to liquids or solids
Say we need 1 pound of a 70% coal tar product, how much of 95% and how much of 50% will we need?
4 = X
grams

34. Example (cont.)
The final product is 1 pound or 454 grams that is represented by 9 total parts (4 + 5)
Solve for X and grams of 95%
= grams of 50%

35. Using Alligation Alternate When the Quantity of One Ingredient Is Known35

36. Example
How many grams of 2.5% hydrocortisone cream should be mixed with 360 g of 0.25% cream to make a 1% hydrocortisone cream?
2.5% part of 2.5% cream 1%
0.25% parts of 0.25% cream
Relative amounts: 0.75:1.5, or 1:2 36

37. Example (cont.)
2 parts = g
1 part X g
X = 180 g (answer) 37

38. Example
How many grams of white petrolatum should be mixed with 250 g of 5% and 750 g of 15% sulfur ointments to prepare a 10% ointment?
12.5% parts of 12.5% mixture
10%
0% parts of white petrolatum
Relative amounts: 10:2.5, or 4:1 38

39. Example (cont.)
4 parts = g
1 part X g
X = 250 g (answer)