How many milliliters of a stock solution of 5.40 M HNO_3 would you have to use to prepare 0.150 L of 0.510 M HNO_3?
(5.40M)*x=(0.150L)(0.510M)
x=0.0142 L
0.0142 L
The problem asks for mL. Convert L to mL.
To solve this problem, we can use the concept of the dilution formula, which is:
C₁V₁ = C₂V₂
Where:
C₁ = initial concentration of the stock solution
V₁ = volume of the stock solution used
C₂ = final concentration desired
V₂ = final volume of the solution
In this case, we are given:
C₁ = 5.40 M (concentration of the stock solution)
V₂ = 0.150 L (final volume of the solution)
C₂ = 0.510 M (final concentration desired)
To find V₁, we rearrange the formula as follows:
V₁ = (C₂ * V₂) / C₁
Substituting the given values, we have:
V₁ = (0.510 M * 0.150 L) / 5.40 M
V₁ = (0.0765 mol/L) / (5.40 mol/L)
V₁ ≈ 0.01416667 L or 14.16667 mL
Therefore, you would need to use approximately 14.16667 mL of the stock solution of 5.40 M HNO₃ to prepare 0.150 L of 0.510 M HNO₃.