A pharmacy receives a medication order for 1 L of a 3% sodium chloride injection. The pharmacy stocks 0.9% sodium chloride injection in 1-L bags and sodium chloride injection, 23.4... A pharmacy receives a medication order for 1 L of a 3% sodium chloride injection. The pharmacy stocks 0.9% sodium chloride injection in 1-L bags and sodium chloride injection, 23.4% in 50-mL vials. How many milliliters of the latter should be added to the 1-L bag of normal saline solution to fill the order?
Understand the Problem
The question is asking us to determine how much of a more concentrated sodium chloride solution (23.4%) is needed to be added to a less concentrated solution (0.9%) to achieve a desired concentration (3%) in a total volume of 1 liter. This involves using the concept of dilution and concentration calculations.
Answer
Approximately \( 93.3 \) mL of the more concentrated sodium chloride solution is needed.
Answer for screen readers
The volume of the more concentrated sodium chloride solution needed is approximately ( 0.0933 ) liters or ( 93.3 ) mL.
Steps to Solve
- Define Variables
Let:
- ( V_1 ) = volume of the more concentrated solution (23.4%)
- ( C_1 ) = concentration of the more concentrated solution = 0.234
- ( V_2 ) = volume of the less concentrated solution (0.9%)
- ( C_2 ) = concentration of the less concentrated solution = 0.009
- ( C_f ) = desired concentration (3%) = 0.03
- ( V_f ) = final total volume = 1 L
- Set Up the Volume Equation
The total volume of the solution after mixing should equal 1 liter:
$$ V_1 + V_2 = 1 $$
- Set Up the Concentration Equation
The total amount of sodium chloride before and after mixing should match. This can be expressed as:
$$ C_1 \cdot V_1 + C_2 \cdot V_2 = C_f \cdot V_f $$
- Substitute the Concentrations into the Equations
Plug the known concentrations into the concentration equation:
$$ 0.234 \cdot V_1 + 0.009 \cdot V_2 = 0.03 \cdot 1 $$
- Express ( V_2 ) in Terms of ( V_1 )
Using the volume equation:
$$ V_2 = 1 - V_1 $$
Plug ( V_2 ) into the concentration equation:
$$ 0.234 \cdot V_1 + 0.009 \cdot (1 - V_1) = 0.03 $$
- Solve the Equation
Now solve for ( V_1 ):
Distributing the terms yields:
$$ 0.234V_1 + 0.009 - 0.009V_1 = 0.03 $$
Combining like terms results in:
$$ 0.225V_1 + 0.009 = 0.03 $$
Subtract 0.009 from both sides:
$$ 0.225V_1 = 0.021 $$
Now divide by 0.225:
$$ V_1 = \frac{0.021}{0.225} $$
Finally, calculate ( V_1 ).
- Calculate ( V_1 )
Perform the final calculation to find ( V_1 ):
$$ V_1 \approx 0.0933 $$ L
Now substitute back to find ( V_2 ):
$$ V_2 = 1 - V_1 \approx 1 - 0.0933 \approx 0.9067 $$ L
The volume of the more concentrated sodium chloride solution needed is approximately ( 0.0933 ) liters or ( 93.3 ) mL.
More Information
To achieve the desired concentration of 3% in a total volume of 1 liter, you'll need about 93.3 mL of the 23.4% solution mixed with roughly 906.7 mL of the 0.9% solution. This is a practical application of dilution which is often used in chemistry, biology, and medicine.
Tips
- Forgetting to convert percentages into decimals when using them in equations. Always ensure to convert concentration percentages into decimal form by dividing by 100.
- Incorrectly assuming that the sum of volumes must equal the sum of concentrations; remember that you need to set equal the total mass of solute before and after mixing.
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