How many ml of .050 M NaOH are required to prepare 1500.0 mL of .0020 M NaOH? Please show work, I'm not sure how to do this. Thanks!
How many moles are in the solution you want to prepare? M x L = moles. So you must add that many moles of the more concentrated solution. L x M = moles again. You know M and you know moles (from the first calculation), solve for L.
Or you can always take the easy way out.
mL x M = mL x M. Plug and chug.
mL x 0.05 = 1500 x 0.002
Solve for mL of the 0.05 M soln.
To answer this question, we can use the concept of the dilution equation. The dilution equation states that the concentration (C₁) and volume (V₁) of a solution before dilution is equal to the concentration (C₂) and volume (V₂) of the solution after dilution. Mathematically, it can be written as:
C₁V₁ = C₂V₂
In this case, we are given:
C₁ = 0.050 M (concentration of the stock solution)
V₁ = unknown (volume of the stock solution needed)
C₂ = 0.0020 M (desired concentration of the diluted solution)
V₂ = 1500.0 mL (volume of the diluted solution)
Plugging these values into the dilution equation, we get:
(0.050 M)(V₁) = (0.0020 M)(1500.0 mL)
To solve for V₁, we rearrange the equation:
V₁ = (0.0020 M)(1500.0 mL) / 0.050 M
Now, let's calculate V₁:
V₁ = (0.0020 M)(1500.0 mL) / 0.050 M
= 60 mL
Therefore, you would need 60 mL of the 0.050 M NaOH solution to prepare 1500.0 mL of the 0.0020 M NaOH solution.