Concentrated nitric acid is 16M. What volume of nitric acid must be diluted with distilled water to prepare 5L of 0.1M nitric acid?
mL x M = mL x M
Can you show me how to work this out. Or give me an example. Please and Thanks
It's just plug and chug.
mL x M = mL x M
mL x 16M = 5,000 mL x 0.1 M
Solve for mL HNO3.
.031
M1V1=M2V2
0.1*5L=16*V2
V2=0.0321 L Dissolve in 5 L of distilled water
To solve this problem, we need to use the formula for dilution:
M1V1 = M2V2
Where:
M1 = initial concentration of the solution to be diluted (in this case, 16M)
V1 = initial volume of the solution to be diluted (what we need to find)
M2 = final concentration of the diluted solution (0.1M)
V2 = final volume of the diluted solution (5L)
So, we can rearrange the formula to solve for V1:
V1 = (M2 * V2) / M1
Substituting the given values into the equation:
V1 = (0.1M * 5L) / 16M
Now, let's calculate the value of V1:
V1 = (0.1 * 5) / 16
= 0.5 / 16
≈ 0.03125 L
Therefore, approximately 0.03125L (or 31.25 mL) of the concentrated nitric acid needs to be diluted with distilled water to prepare 5L of 0.1M nitric acid.