Order Entry and Processing Study Guide for the PTCB Exam
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Vocabulary and Procedures
As a pharmacy technician, it is important to know and understand pharmacy related verbiage, calculations, and procedures. You must be able to “talk the talk” to be able to “walk the walk” as they say. Being competent in these areas will help with effective communication, patient care, and ultimately success in your daily tasks and duties.
Calculations
A pharmacy technician must have a firm understanding of basic pharmacy calculations, regardless of whether their specific daily tasks involve such calculations or not. Although the pharmacist-on-duty will ultimately double check that calculations are done correctly, it makes for a more efficient and fluid workflow if the pharmacy technician is proficient and confident in their mathematical skills.
Formulas
An important method to know and use when performing pharmacy calculation is called dimensional analysis. This method uses ratios in a sequenced formula to calculate a wanted quantity or answer. The general dimensional analysis blueprint is as follows:
(Given QT) \(\times\) (CF) = (Wanted QT)
QT is the abbreviation for quantity
CF is the abbreviation for conversion factor
Example: Change 20 inches to cm.
Use \(1 \ in = 2.54 \ cm\) to make a conversion factor, CF, and multiply by the given quantity, QT.
\[20 \ in \times \dfrac{2.54 \ cm}{1 \ in} = 50.8 \ cm\]You may need two conversion factors in some problems. (Given QT) \(\times\) (CF) \(\times\) (CF)= (Wanted QT)
For example, JT is a male patient who weighs 141lb. His doctor wants to prescribe a medication that has a specific weight base dosing of 5 mg/kg per day. How many mg of medication should be prescribed per day?
Since 1 kg = 2.2 lb, the first conversion factor will be \(\frac{1 \ kg}{2.2 \ lb}\) to convert lb of weight to kg.
Since the dose rate is \(\frac{5 \ mg}{1 \ kg}\), that will be used as the second conversion factor to convert kg of body mass into mg of medication.
\[141 \ lb \times \frac{1 \ kg}{2.2 \ lb} \times \dfrac{5 \ mg}{1 \ kg}\]After crossing out like units, we have
\[\dfrac{141 \times 1 \times 5 \ mg}{2.2 \times 1 } = 320.45 \ mg\]Ratios and Proportions
A ratio can be summarized as the relative value between two numbers. A fraction of \(\frac{1}{2}\) could be read as a ratio of 1:2 or one part is to two parts. Magic mouthwash was mentioned above in the Mixtures subsection with a parts per ratio of 1:1:1. This ratio shows that each ingredient of the mixture has equal parts. A fraction of \(\frac{4}{8}\) can be reduced down and equal the fraction of \(\frac{2}{4}\) and furthermore to \(\frac{1}{2}\), with all three equaling the ratio 1:2. This means that they all are equivalent.
A proportion expresses the equality of two ratios. Proportions can be seen in a few different ways, which include:
- a:b = c:d
- a/b = c/d
- a:b :: c:d
Setting up a proportion helps to find the missing value in the equivalent expressions. For example, if 100 mg of morphine concentrate equals 5 mL, then how many mLs are needed to provide 10 mg?
\[\dfrac{100 \ mg}{10 \ mg} = \dfrac{5 mL}{x}\] \[100 \ x = 50 \ mL\]which can be expressed as:
\[x=\dfrac{50 \ mL}{100}\]resulting in:
\[x = 0.5 \ mL\]So, \(\dfrac{100 \ mg}{5 \ mL}\) is equivalent to \(\dfrac{10 \ mg}{0.5 mL}\), with both equaling the ratio 20:1.
Alligations
Another type of calculation commonly seen is the alligation method, which allows you to prepare a concentration of solution from two other solutions of the same ingredients but in different strengths. One solution needs to be a higher strength and the other a lower strength than the desired concentration of the final product. Using the differences in % strengths, the number of parts of each can be determined, and the volume of each part can be calculated based on the total solution volume of the desired product.
See the example calculations below:
- Number of parts for higher % solution = Desired % in final product - Lower % solution Answer = C - B
- Number of parts for lower % solution = Higher % solution - Desired % in final product Answer = A - C
- Add the number of parts together to get a total number of parts.
- Divide the final volume of the desired product by the total number of parts to get mL/part.
- Multiply the number of parts for each % solution by the mL/part.
To get the volume of each solution, you need to combine [how much of the high % and low % solution to add, respectively] to get the desired final product.
Example:
How many mLs of a 60% solution and a 25% solution are needed to make 700 mL of a 30% solution?
- Number of parts for higher % solution = Desired % in final product - Lower % solution
Answer: 30 - 25 = 5 - Number of parts for lower % solution = Higher % solution - Desired % in final product
Answer: 60 - 30 = 30 - Add the number of parts together to get a total number of parts.
Answer: 30 + 5 = 35 total parts - Divide the final volume of the desired product by the total number of parts to get mL/part.
Answer: 700/35 parts = 20 mL/part - Multiply the number of parts for each % solution by the mL/part.
To get the volume of each solution, you need to combine [how much of the high % and low % solution to add, respectively] to get the desired final product.
Answer a: 20 mL/part x 5 parts = 100 mL of 60% solution
Answer b: 20 mL/part x 30 = 600 mL of 25% solution
Conversions
As stated in previous modules, it is important to know units of measure and how to convert from one to another. Here is a simple list of different units of measure conversions that should be committed to memory:
| Volume | Weight | Length |
|---|---|---|
| 1 tsp = 5 mL | 1 kg = 2.2 lb | 1 inch = 2.54 cm |
| 1 tbsp = 15 mL | 1 gm = 10 dg = 100 cg = 1000 mg = 1 000 000 mcg |
|
| 1 L = 10 dL = 100 cL = 1000 mL = 1 000 000 mcL |
1 grain = 64.8 mg | |
| 1 mL = ~20 drops | 1 lb = 16 oz | |
| 1 gal = 4 qts | 1 lb = ~454 gm | |
| 1 fl oz = 29.6 mL | ||
| 1 qt = 2 pints | ||
| 1 pt = ~473 mL |
Knowing these will help when performing dimensional analysis, when calculating dosing, or converting from one dosing unit of measure to another.
For example, a patient was given a prescription for eye drop X, with the directions of: 1 drop in each eye every 4 hours for 7 days. Quantity: QS. Eye drop X is commercially available in the following quantities: 2.5 mL, 5 mL, 10 mL, and 20 mL. Which would you dispense to be sufficient for the patient’s course of therapy? 1 drop in each eye every 4 hours for 7 days equals 84 total drops. 84 drops/20 drops = 4.2 mL. A 5 mL bottle is the smallest package size and best reasonable choice to dispense to sufficiently cover this patient’s 7-day course of therapy.
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