How to Calculate the Volume of Nitric Acid Needed for Dilution?
What volume of 15.9 M Nitric acid would be required to make 3.2 L of 2.00 M nitric acid?
Final answer: To make 3.2 L of 2.00 M nitric acid from 15.9 M nitric acid, you will need 0.402 L of the 15.9 M nitric acid.
Answer:
To calculate the volume of 15.9 M nitric acid required to make 3.2 L of 2.00 M nitric acid, we can use the equation:
(M1)(V1) = (M2)(V2)
Where M1 and V1 are the molarity and volume of the stock solution, and M2 and V2 are the desired molarity and volume of the final solution.
Given:
M1 = 15.9 M
V2 = 3.2 L
M2 = 2.00 M
Substituting these values into the equation:
(15.9 M)(V1) = (2.00 M)(3.2 L)
Solving for V1:
V1 = (2.00 M)(3.2 L) / 15.9 M = 0.402 L
Therefore, 0.402 L of the 15.9 M Nitric acid is required to make 3.2 L of 2.00 M nitric acid.
When diluting a concentrated solution such as 15.9 M Nitric acid to a desired concentration like 2.00 M, it is important to calculate the volume of the concentrated solution needed.
In this case, we used the equation (M1)(V1) = (M2)(V2) where M1 represents the molarity of the stock solution, V1 is the volume of the stock solution needed, M2 is the desired molarity of the final solution, and V2 is the final volume of the diluted solution.
By substituting the given values of M1, V2, and M2 into the equation, we were able to calculate that 0.402 L of the 15.9 M Nitric acid is required to make 3.2 L of 2.00 M nitric acid. This calculation ensures that the desired concentration of nitric acid is achieved accurately.