You wish to prepare 0.18 M HNO3 from a stock solution of nitric acid that is 18.9 M. How many milliliters of the stock solution do you require to make up 1.00 L of 0.18 M HNO3?
M x mL = M x mL.
18.9 M x mL = 0.18 M x 1000 mL
I don't understand what to do.
solve for ml
(it's a dilution equation)
You have the initial concentration and final concentration with volume of it that you want. All you need is the initial volume of concentrated nitric acid you want to remove from the concentrated stock volume, to use to dilute to the needed concentration.
To solve this problem, we can use the formula:
C1V1 = C2V2
Where:
C1 = concentration of the stock solution (18.9 M)
V1 = volume of the stock solution (unknown)
C2 = concentration of the final solution (0.18 M)
V2 = volume of the final solution (1.00 L)
We need to find V1, the volume of the stock solution required to make up 1.00 L of 0.18 M HNO3.
Rearranging the formula, we have:
V1 = (C2V2) / C1
Let's plug in the values:
C1 = 18.9 M
V2 = 1.00 L
C2 = 0.18 M
V1 = (0.18 M * 1.00 L) / 18.9 M
Now we can perform the calculation:
V1 = 0.00952 L = 9.52 mL
Therefore, you would require 9.52 mL of the stock solution to make up 1.00 L of 0.18 M HNO3 solution.