Use the dilution equation: M i V i = M f V f .
Rearrange the equation to solve for the initial volume: V i = M i M f V f .
Plug in the given values: V i = 4.00 M ( 0.50 M ) ( 12.0 mL ) .
Calculate the initial volume: V i = 1.5 mL . The required volume is 1.5 mL .
Explanation
Problem Analysis We are given a problem involving the dilution of a sulfuric acid solution. We have an initial concentration of sulfuric acid ( H 2 S O 4 ) of 4.00 M, and we want to prepare 12.0 mL of a 0.50 M H 2 S O 4 solution. Our goal is to find the volume of the initial solution needed for this dilution.
Dilution Equation To solve this problem, we will use the dilution equation, which states that the product of the initial concentration and volume is equal to the product of the final concentration and volume: M i V i = M f V f where:
M i is the initial concentration
V i is the initial volume (what we want to find)
M f is the final concentration
V f is the final volume
Rearranging the Equation We need to solve for V i , so we rearrange the equation: V i = M i M f V f
Plugging in the Values Now, we plug in the given values:
M i = 4.00 M
M f = 0.50 M
V f = 12.0 mL V i = 4.00 M ( 0.50 M ) ( 12.0 mL )
Calculating the Volume Calculating the initial volume: V i = 4.00 0.50 × 12.0 = 4.00 6.0 = 1.5 mL
Final Answer Therefore, you would need 1.5 mL of the 4.00 M sulfuric acid solution to make 12.0 mL of a 0.50 M sulfuric acid solution.
Examples
Dilution calculations are commonly used in chemistry, biology, and medicine. For example, when preparing medication, a pharmacist needs to dilute a concentrated stock solution to achieve the correct dosage for a patient. Similarly, in a laboratory, researchers dilute stock solutions of chemicals to create solutions with specific concentrations for experiments. Understanding dilution calculations ensures accuracy and safety in these applications.
To prepare a 0.50 M solution of sulfuric acid, you need to use 1.5 mL of the 4.00 M solution. This was calculated using the dilution equation: V i = M i M f V f . By substituting in the known values, we found that the required initial volume is 1.5 mL.
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