Calculate how to prepare 100.0mL of a 1.25x10^-4 M aqueous solution of Fe(NO3)3(aq) by dilution of stock of 0.0025M Fe(NO3)3(aq) in a 100.0mL volumetric flask?

I tried using the m1v1=m2v2 equation but it doesn't look like that equation would work for this problem??

You're right, the M1V1 = M2V2 equation doesn't directly apply to this problem. In this case, you will need to calculate the dilution using the concept of moles.

To prepare the desired solution:
1. Determine the moles of Fe(NO3)3 needed in the final solution:
- Moles = Molarity x Volume
- Moles = (1.25x10^-4 mol/L) x (0.100 L)
- Moles = 1.25x10^-5 mol

2. Calculate the volume of stock solution required:
- Molarity(stock) x Volume(stock) = Moles
- Volume(stock) = Moles / Molarity(stock)
- Volume(stock) = (1.25x10^-5 mol) / (0.0025 mol/L)
- Volume(stock) = 0.005 L or 5.0 mL

3. Add the calculated volume of stock solution (5.0 mL) to the 100.0 mL volumetric flask.

4. Add distilled water to the flask until the final volume reaches 100.0 mL, ensuring thorough mixing.

By following these steps, you will prepare 100.0 mL of a 1.25x10^-4 M aqueous solution of Fe(NO3)3.

You're right. The equation you mentioned, m1v1 = m2v2, is commonly used for dilution problems, but in this case, it won't be useful because we are not given the volumes of the stock solution and the final solution. Instead, we are given a 100.0 mL volumetric flask, which tells us the final volume of the solution.

To solve this problem, we can use the concept of moles and the formula:

Molarity = moles / volume

Here's how you can calculate the required amount of stock solution to prepare the desired final solution:

1. Calculate the number of moles of Fe(NO3)3 needed in the final solution:
Moles = Molarity x Volume
Moles = (1.25 x 10^-4 mol/L) x (0.100 L)
Moles = 1.25 x 10^-5 mol

2. Now, we'll calculate the volume of the stock solution required to obtain the desired number of moles. We can use the same formula as before but rearrange it:
Volume = Moles / Molarity
Volume = (1.25 x 10^-5 mol) / (0.0025 mol/L)
Volume = 5 mL

So, you'll need to measure 5 mL of the stock solution (0.0025 M Fe(NO3)3(aq)) using a pipette or a graduated cylinder and then transfer it to the 100.0 mL volumetric flask. After that, fill the flask up to the mark with distilled water to reach a final volume of 100.0 mL. This will give you the desired 1.25 x 10^-4 M aqueous solution of Fe(NO3)3(aq).

Don't know why.

1.25E-4M x 100 mL = 0.0025M*mL
Solve for mL of the 0.0025 M solution, add that to a 100 mL volumetric flask, dilute to the mark, mix thoroughly, stopper, voila!