Pharmaceutical Calculations for the Pharmacy Technician Copyright © 2008 Wolters Kluwer Health |...

Pharmaceutical Calculations for the Pharmacy Technician Copyright © 2008 Wolters Kluwer Health | Lippincott Williams & Wilkins

Chapter 10

Dilution and Concentration


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


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


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


Definitions

• Diluent:– 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


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


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.)


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


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


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)


Example

• If 50 mL of a 1:20 (w/v) solution of aluminum acetate are diluted to 1000 mL, what is the ratio strength (w/v)?

• Use same formula

(50 mL) (5 %) = 1000 mL (X %)

X = 0.25% = 1:400 (answer)


Determine Amount of Solution of a Desired Strength• How many grams of 10% (w/w)

ammonia solution can be made from 1800 g of 28% (w/w) strong ammonia solution?

• Use the same formula

(1800 g)(28 %) = (X g) (10%)

X = 5040 g (answer)


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%


(125 mL) (0.2 %) =(X mL) (0.02 %)

X = 1250 mL (answer)


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% 1:2000 = 0.05%


X mL (0.25 %) = 4000 mL (0.05%)

X = 800 mL (answer)


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% 1:2000 = 0.05%

X (0.25%) = (3785 mL)(0.05%)

X = 757 mL (answer)


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?


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


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


In order to use our formula we must calculate the percent strength of each ingredient5 mg/mL = 0.005 g/mL which is a 0.5% solution65 mg/mL = 0.065 g/mL = 6.5% solutionX (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 = 0.005 g/mL which is a 0.5% solution

• 65 mg/mL = 0.065 g/mL = 6.5% solution

X (6.5%) = (30 mL) (0.5%)

X = 2.3 ml of phenobarbital required (answer)


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


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


Example (cont.)

5 mL = 0.5 g 50 mL X g

X = 5 g (answer)


Amount of Diluent Needed for Preparing Solution of Specified Lower Strength


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 = 997.5 or 1000 mL


Example (cont.)

– 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)


Alligation

• Arithmetical method of solving problems that involves mixing of solutions or solids that have different percentage strengths


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


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


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


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


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


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 9 454 grams


Example (cont.)

– The final product is 1 pound or 454 grams that is represented by 9 total parts (4 + 5)

– Solve for X and 201.7 grams of 95%

– 454 - 201.7 = 252.3 grams of 50%


Using Alligation Alternate When the Quantity of One Ingredient Is Known


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% 0.75 part of 2.5% cream

1%

0.25% 1.5 parts of 0.25% cream

Relative amounts: 0.75:1.5, or 1:2


Example (cont.)

2 parts = 360 g 1 part X g

X = 180 g (answer)


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% 10 parts of 12.5% mixture

10%

0% 2.5 parts of white petrolatum

Relative amounts: 10:2.5, or 4:1


Example (cont.)

4 parts = 1000 g 1 part X g

X = 250 g (answer)

Pharmaceutical Calculations for the Pharmacy Technician Copyright © 2008 Wolters Kluwer Health |...

Documents

Transcript of Pharmaceutical Calculations for the Pharmacy Technician Copyright © 2008 Wolters Kluwer Health |...