By the end of this webpage, you will be able to:
define the expressions of concentrations namely percentage weight-in-volume, percentage weight-in weight, percentage volume-in-volume.
define the expression ratio strength in its simplest form, milligram percent, parts-per-million, and parts-per-billion.
convert percentage strength expressions to ratio strength expressions.
express concentration of mixtures in percentage, ratio strength and milligram percent.
solve application problems involving percent strength, ratio strength, weight-in-volume, weight-in-weight, volume-in-volume, milligram percent for pharmaceutical preparations with 100% accuracy without assistance.
In pharmacy, concentration refers to the amount of a substance (usually a drug) present in a given volume of a solution. It is a measure of the strength or potency of a solution and is crucial for accurate dosing and preparation of medications.
The definition of concentration used in pharmacy practice is indeed closely aligned with the guiding principles in chemistry from which the concept of concentration is derived. In chemistry, concentration refers to the amount of a constituent or substance (which can be a solid, liquid, or gas) divided by the total volume of a mixture. The basic units of measurement are grams and milliliters.
Concentration = Amount of Substance (grams)/Total Volume (mL)
Concentration may be expressed in many ways. Click on each tab to view the related descriptions.
Concentrations can be expressed as a percentage or ratio of the total solution. For example, a 2% solution means there are 2 grams of the active ingredient in 100 milliliters of the solution.
Percentage weight-in-volume (% w/v) represents the grams of a solid substance per 100 mL of a preparation (such as suspension or solution) and is a variation of the percentage strength expression. Other forms include percentage volume-in-volume (% v/v), indicating milliliters of a substance per 100 mL of preparation, and percentage weight-in-weight (% w/w), denoting grams of a substance per 100 grams of preparation.
Let us look at some application questions:
Example 1
If one is to prepare 120 mL of a 5% w/v Acetaminophen suspension using Acetaminophen 500 mg tablets and a suitable vehicle, what would be the total amount of tablets needed?
Step 1- Determine the quantity of Acetaminophen in the final product.
The concentration of the final product should be 5% w/v; meaning, there is 5 g of Acetaminophen for every 100 mL of preparation.
Therefore the final product should contain: (5 g * 120 mL)/100 mL = 6 g
Step 2- Determine how many tablets are needed to make the suspension.
convert mg → grams = 500 mg/1,000 = 0.5 g = 1 Acetaminophen tablet.
If, 1 tablet/0.5 g = x tablets/6 g = (1 tablet * 6 g)/0.5 g = 12 tablets.
Step 3- Document the Compounding Formula
New Formula- Acetaminophen Suspension 5% w/v
Acetaminophen Tablets 500 mg 12 tablets
Suitable suspension vehicle q.s. 120 mL
Example 2
A 2% w/v stock concentration of Diphenhydramine was diluted to obtain 6 bottles of Diphenhydramine 12.5 mg/5 mL each containing 120 mL. What is the volume of Diphenhydramine concentrate used in this compounding scenario?
Step 1- Determine the quantity of Diphenhydramine in each bottle.
If, the concentration is 12.5 mg in 5 mL, there will be x mg in 120 mL
x mg = (12.5 mg * 120 mL)/5 mL = 300 mg of Diphenhydramine in each bottle.
Step 2- Determine the amount of Diphenhydramine needed to compound 6 bottles.
= 300 mg/bottle * 6 bottles = 1,800 mg
Step 3- Determine the volume of stock that will yield 720 mg of Diphenhydramine to compound the the 6 bottles.
If, the stock concentration is 2% then there are 2 g (2,000 mg) in each 100 mL
Therefore, 1,800 mg will be in x mL of stock.
x mL = (1,800 mg * 100 mL)/2,000 mg = 90 mL
Step 4- Document the Compounding Formula
New Formula- Diphenhydramine 12.5 mg/5 mL
Diphenhydramine 2% w/v 90 mL
Suitable diluent ad. 720 mL
Ratios are used to represent the proportion of the active ingredient to the total solution. For example, a ratio of 1:100 means 1 part of the active ingredient is present in 100 parts of the solution.
Ratio states the relationship between one quantity and another and may be written as a common fraction or with a colon between the two numbers. For example, three parts compared with four parts may be written as 3:4 or ¾. The amount of the substance in the preparation expressed as a ratio is referred to as ratio strength.
We can convert between percentage strength and ratio strength interchangeably. If a product’s concentration is given in ratio strength and specific calculations are required, it’s easy to convert between different expressions. For instance, a stock concentration of 20% w/v can be converted to ratio strength as follows:
20% w/v means 20 g of active substance in 100 mL of preparation.
In other words, 20 g/100 mL
Ratio strength should be expressed in its simplest form, meaning one should readily see the numerical relationship between 1 part of the active substance and the whole. Dividing the numerator and denominator by the highest common multiple may be considered; however, this does not always simplify the proportion. It is best to divide the numerator by itself and then divide the denominator by the initial value in the numerator. Let’s see what this means in the example.
20 g/100 mL when simplified = 20/20 : 100/20 = 1 : 5 w/v
Recall, we started with 20% w/v and so in ratio terms it is 1 part of active substance in grams to every 5 parts of preparation in mL.
Another Example- Converting 33% v/v to ratio strength
= 33 mL/100 mL
= 33/33 : 100/33
= 1 : 3.03 v/v meaning, 1 part of active substance in mL to every 3.03 parts of preparation in mL
In general, the strength of a substance in a liquid is typically expressed as the number of grams per 100 mL. However, there are cases where substances are denoted in milligrams percent (mg%). This notation is commonly utilized to represent concentrations of natural substances in biological fluids, indicating the number of milligrams per 100 mL of the solution.
An example: results of the blood urea nitrogen (BUN) test are measured in milligrams per deciliter (mg/dL). It is important to note that 1 dL equals 100 mL. Therefore, if the normal range for BUN is 6-24 mg/dL, this may also be represented as 6-24 mg%.
Example
A medication is prescribed to be administered intravenously at a concentration of 5 mg% for a patient weighing 70 kg. Calculate the total amount of medication (in milligrams) to be administered to the patient for a single dose.
5 mg% means 5 mg of active ingredients in each 100 mL of preparation.
In each single dose, the patient will receive 5 mg of the drug.
These notations, expressed as parts per million (ppm) or parts per billion (ppb), are utilized to represent minuscule concentrations, denoting the number of parts of solute per million or billion parts of the solution, respectively. This quantifies the presence of an agent within one million or billion parts of a mixture.
For instance, fluorinated drinking water often contains one part of fluorine per million parts of drinking water, which is numerically expressed as 1:1,000,000. The unit of weight measurement is the gram (g), while volume is measured in milliliters (mL).
The notation 5 ppm means 5 parts active per 1,000,000 parts of preparation. Likewise, 6 ppb means 6 parts of active per 1,000,000,000 parts of preparation. These expressions may be further simplified:
5 ppm = 5 : 1,000,000 = 5/5 : 1,000,000/5 = 1 : 200,000 when simplified.
and 6 ppb = 6 : 1,000,000,000 = 6/6 : 1,000,000,000/6 = 1 : 166,666,666.7 when simplified.
These will not be covered on this page, however, it is good to be aware of these other expressions:
Mass/Volume Ratio– This notation represents the mass of the solute (drug) in a given volume of solution (usually milligrams per milliliter or grams per liter).
Volume/Volume Ratio- Concentrations can be expressed as volumes of solute in a given volume of solution (milliliters per liter).
Molarity (M)- Molarity represents the number of moles of solute per liter of solution. It provides a measure of the number of chemical entities (molecules or ions) in a given volume of solution.
Milliequivalents per Liter (mEq/L)– This notation is commonly used for electrolytes and represents the number of milliequivalents of ions per liter of solution
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Dilution, Concentration and Alligation
By the end of this section you should be able to:
- differentiate between dilution and concentration of a pharmaceutical preparation
- explain special considerations for altering product strength in pharmaceutical compounding
- perform calculations for altering product strength by dilution using the concept of ratio and proportion with 100% accuracy.
- apply the equation CsVs = CdVd with 100% accuracy and without assistance.
- perform calculations for altering product strength by dilution using the equation CsVs=CdVd with 100% accuracy and without assistance
- perform calculations for preparing a stock preparation with 100% accuracy and without assistance
- perform calculations for reconstituting powders for suspensions and solution using ratio and proportion and CsVs=CdVd equation with 100% accuracy and without assistance
- perform calculations for altering (fortifying or diluting) the concentration of a given formula with 100% accuracy
Altering Product Strength
The strength of a preparation signifies the concentration of the mixture. This concentration can be adjusted by either increasing or decreasing it. Typically, we dilute portions of the stock to create a less concentrated mixture by adding a diluent. Conversely, concentration can be enhanced, making it more potent, by adding more of the active ingredient; this is known as fortification.
Let us explore how we can do this using the following methods. Click on each tab below to review the related content.
Understanding Dosage Forms
In our previous lesson, we emphasized that a dosage form is more than just the pure drug; it incorporates the active pharmaceutical ingredient (API), responsible for therapeutic effects, along with inactive components known as excipients. When adjusting concentration, it’s crucial to consider the proportion of these excipients in relationship to the proportion of the active ingredient. In essence, this reflects how much of the active substance is present per specific volume of the entire product. This ratio is fundamental to dosage form preparation, ensuring the right balance between therapeutic efficacy and safe administration.
Stock Preparation Essentials
Compounding a stock preparation involves several key components. Firstly, the pure drug or API, which can exist in various forms such as solids or liquids. The vehicle or solvent, serving as the base, might be a liquid, semi-solid, or solid depending on the specific formulation requirements. Typically, the solute is soluble in the solvent, resulting in a stock solution. However, there are instances where the stock preparation is in the form of a suspension, solid (like powders, tablets, or capsules), or a semi-solid like creams, gels, pastes, or pessaries. Each formulation type demands careful consideration of the physical characteristics of both the API and the base preparation if one must dilute or fortify the preparation.
Factors Influencing Dilution or Concentration
Before any adjustments are made to the product, several factors must be taken into account. The type of formulation, whether it’s an ointment, cream, dry powder, solution, gel, paste, etc., plays a pivotal role in determining the concentration alterations. Additionally, understanding the physical properties of the pure drug and the base preparation is essential. Moreover, patient-specific parameters, including age, gender, and existing disease states, need thorough evaluation. These considerations ensure that any modifications to concentration align not only with the formulation’s integrity but also with the individual patient’s needs and safety.
Let us look at some examples. We will prepare a stock solution of sodium chloride (NaCl) and then use this stock solution to make other diluted preparations.
If one is to prepare 1 liter of a 15% w/v sodium chloride stock solution using sodium chloride powder and water, determine how many grams of the salt will be required.
In this scenario, the stock comprises NaCl powder and purified water. The NaCl powder is considered as pure NaCl, presenting a concentration of 100% w/w. Conversely, purified water is denoted as having a concentration of 0%; this is because it acts as the diluent in the mixture. According to the concentration definition (Amount of substance/Volume), purified water contains no NaCl, thus resulting in 0% w/v.
This means that 150 g of the NaCl will be dissolved in a portion of water. Once completely dissolved, the mixture will be made up to 1,000 mL with purified water.
Now that we prepared our stock solution let us make some other diluted products from this NaCl 15% w/v stock solution.
If one is to prepare 1 liter of a 0.9% w/v isotonic solution, determine how many milliliters of the stock solution 15% w/v sodium chloride solution would be needed to prepare the isotonic solution.
Step 1- What do I have/want?
Have- NaCl 15 % w/v stock solution
Want- 1,000 mL of NaCl 0.9% w/v Solution.
Step 2- Determine the amount of API contained in the product.
0.9% w/v means, 0.9 g in 100 mL
therefore, x g will be needed in 1,000 mL
X = (0.9 g * 1,000 mL)/100 mL = 9 gram of NaCl.
Question: Where will I find the 9 g of NaCl to compound the solution?
Step 3- Determine the volume of the stock solution which will contain 9 g of NaCl.
15% w/v means, 15 g in 100 mL
Therefore, 9 g will be in x mL of the stock.
X = (9 g * 100 mL)/15 g = 60 mL of stock solution.
Step 4- Write the New Formula.
NaCl 0.9% w/v Solution
NaCl 15% w/v solution 60mL
Purified water ad. 1,000 mL
Next task- You are given the master formula below to prepare a 0.9% w/v isotonic solution. Determine the quantities of ingredients needed to prepare 500 mL of the solution. Stocks available include: Sterile water for injection and 15% w/v sodium chloride stock solution.
Next task- You are given the master formula below to prepare a 0.45% w/v hypotonic solution. Determine the quantities of ingredients needed to prepare 500 mL of the solution. Stocks available include: Sterile water for injection and 15% w/v sodium chloride stock solution.
Master Formula for 0.45% w/v Solution
NaCl 15% w/v solution 30 mL
Purified water q.s. 1000 mL
Here will reduce the formula volume while maintaining the same concentration.
Vs = 0.45 x 500/15
= 15 mL of 15% w/v NaCl stock solution.
New Formula for 0.45% w/v Solution
NaCl 15% w/v solution 15 mL
Purified Water q.s. 500 mL
Did you notice any difference with the last two calculations? If you did, Great!
In the last two examples we used the equation,
CsVs = CdVd
to solve for the volume of stock (Vs) to be used to make up the product. If you used ratio and proportion to solve the problem your responses should be the same in both examples as well.
Click on the other tab to explore CsVs = CdVd in greater details.
Many drugs, especially antibiotics, tend to lose their potency rapidly when prepared as liquid dosage forms. To address this, these drugs are formulated as dry powders. Upon dispensing, these powders must be reconstituted with a diluent. The reconstituted liquid typically maintains stability for up to 14 days under refrigeration.
In the case of suspensions for oral or parenteral use, they consist of finely divided drug particles dispersed in a liquid vehicle. Solutions for oral or parenteral use, on the other hand, contain one or more solutes dissolved in a solvent or a mixture of solvents. A diluent is utilized as a vehicle to facilitate the dispersion of insoluble powders or the dissolution of soluble powders.
Reconstituted powders are obtained by transforming the original liquid form into powder, achieved by removing the solvent. To restore the formulation to its original liquid state, a specific quantity of liquid is added. This reconstitution process ensures the drug’s stability and efficacy, allowing for accurate and safe administration.
In order to perform operations using the CsVs=CdVd equation, there are four key parameters that must be correctly identified. It is customary for you to receive three of the four parameters and use your technical expertise to determine the value of the unknown parameter.
Cs = Concentration of the stock (we looked at the various expressions of concentration in previous content and so that knowledge will be useful in correctly identifying the information).
Vs = Volume of the stock (you may need to use the entire volume provided to make the desired product or you maybe asked to determine the volume to be used to prepare the product/dilution).
Cd = concentration of the dilution (kindly note the expression of concentration on both sides of the equation must be the same).
Vd = volume of dilution (this is the total volume of the product you are making. Note carefully the volume of the dilution is not the same as the volume of diluent to be added to a product. The volume of the dilution is the sum of the volume of API and Volume of the diluent).
Let us look at the following example.
Rx
Cephalexin 187 mg po tid * 1/52
Cephalexin is available as 125 mg/5 mL and 250 mg/5 mL. Select the appropriate propriate product to fill the prescription such that the dose 187 mg be in 5 mL of reconstituted suspension.
Step 1- Determine the Stock to use for this prescription.
187 mg/5 mL > 125 mg/5 mL- diluting the product will further reduce the concentration. The bottle may also contain insufficient amount of antibiotic for the prescription.
250 mg/5 mL is > than 187 mg/5 mL- therefore, it can be diluted to a lower concentration. One bottle of this suspension may be sufficient to fill the prescription order.
Step 2- Determine the parameters for the CsVs=CdVd equation.
Cs = 250 mg/5 mL (5% w/v)
Vs = 100 mL
Cd = 187 mg/5 mL (3.74% w/v)
Vd = ?
Step 3- Determine the unknown, Vd.
Now, recall Vd is the total volume of the dilution and so we need to determine the volume of water for reconstitution needed such that the suspension when reconstituted to 133.7 mL will yield a concentration of 187 mg/5 mL.
Step 4- Determine the volume of powder in the stock to be used.
Volume of Powder = Volume of stock (80 mL) – Volume of water for reconstitution (70 mL) = 30 mL
Step 5- Determine volume of water for reconstitution needed to mix the antibiotic.
Water for reconstitution = Volume of dilution (133.7 mL) – Volume of Powder (30 mL) = 103.7 mL
Lets work one more example.
You are asked to supply alcohol 1 : 1.4286 v/v to an out-patient clinic. The supplies are late however, you have 95% v/v Alcohol and water for irrigation. Show calculations to prepare 1,500 mL of alcohol for the out-patient clinic.
Step 1- Determine the Parameters for the equation.
Cs= 95%
Vs = ?
Cd = 1 : 1.4286
Vd = 1,500 mL
The New Product Formula
Alcohol 95% v/v 1,105.3 mL
Water for irrigation ad. 1,500 mL
The recoding below uses ratio and proportion and CsVs=CdVd interchangeably in altering product concentration.
Alligation Medial
This method enables the effortless calculation of the “weighted average” or percentage strength for a mixture comprising two or more substances with known quantities and concentrations. To use this method you must know the concentration as well as the volume of each stock added to the mixture. It is mostly used for liquid but may also be used for semisolids such as: ointment, creams, gels, etc.
Alligation Alternate
A mathematical technique employed to determine the quantity of different components or mixtures with known strengths (concentrations) needed to create a desired product of specific strength and quantity. It’s important to note that in this method, the proportions in which each of the stocks must be mixed are initially unknown and must be calculated.
Lets review the recording below to better understand these concepts.
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By: D. Baker (BPharm, Dip.Ed.)
Published: 2023- 11- 10, Last updated: 2024- 04- 07.