How would a student make 500.0 mL of a 0.08500 M solution of iron(III) using a stock solution that is 0.3180 M in iron(III)? Explain how this would be done.

I know not how a student would do it but I would use c1v1 = c2v2

0.08500M*500 mL = 0.3180M*v2
v2 = 133.6 mL
The way you do this is to pipet or measure accurately 133.6 mL of the stock solution and transfer quantitatively to a 500 mL volumetric flask. Then add distilled water to the mark of the volumetric flask.

To make a 500.0 mL solution of 0.08500 M iron(III) using a stock solution of 0.3180 M iron(III), you need to follow a step-by-step process. Here's how it can be done:

Step 1: Calculate the amount of iron(III) needed.
To determine the amount of iron(III) required to make the desired solution, you can use the formula:

C₁V₁ = C₂V₂

Where C₁ is the concentration of the stock solution (0.3180 M), V₁ is the volume of the stock solution used (unknown), C₂ is the concentration of the desired solution (0.08500 M), and V₂ is the final volume of the desired solution (500.0 mL or 0.5000 L).

To find V₁, you rearrange the equation:

V₁ = (C₂ * V₂) / C₁

Substituting the values:

V₁ = (0.08500 M * 0.5000 L) / 0.3180 M

Simplifying gives:

V₁ ≈ 0.1336 L or 133.6 mL

Therefore, you need approximately 133.6 mL of the stock solution.

Step 2: Add the stock solution to a container.
Measure out 133.6 mL of the 0.3180 M iron(III) stock solution using a graduated cylinder or pipette.

Step 3: Dilute with distilled water.
After measuring the stock solution, transfer it to a larger container, such as a volumetric flask or beaker. Then, add enough distilled water to bring the total volume up to 500.0 mL.

Step 4: Mix well.
Thoroughly mix the solution by swirling or using a stirring rod to ensure the iron(III) ions are evenly distributed throughout the solution.

By following these steps, a student can make a 500.0 mL solution of 0.08500 M iron(III) using a stock solution with a concentration of 0.3180 M.

To make a 500.0 mL solution of 0.08500 M iron(III) using a 0.3180 M iron(III) stock solution, you would need to dilute the stock solution with an appropriate volume of water. Here's how you can do it step-by-step:

Step 1: Determine the volume of the stock solution required:
The formula for dilution is C1V1 = C2V2, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution, C2 is the desired concentration of the final solution, and V2 is the desired volume of the final solution.

Given:
C1 = 0.3180 M (concentration of stock solution)
C2 = 0.08500 M (desired concentration of final solution)
V2 = 500.0 mL (desired volume of final solution)

Rearranging the formula, we get:
V1 = (C2 x V2) / C1

Substituting the values:
V1 = (0.08500 M x 500.0 mL) / 0.3180 M
V1 = 133.647 mL (rounded to three decimal places)

So, you would need 133.647 mL of the stock solution to prepare the desired solution.

Step 2: Measure the required volume of the stock solution:
Using a graduated cylinder or a pipette, carefully measure 133.647 mL of the 0.3180 M iron(III) stock solution. Be accurate while measuring to ensure the desired concentration is achieved.

Step 3: Dilute with water:
After measuring the stock solution, transfer it to a volumetric flask or a container that can hold the desired final volume (500.0 mL in this case). Then, add water to the flask until the total volume reaches 500.0 mL, ensuring thorough mixing to obtain a homogeneous solution.

Step 4: Label and store:
Label the flask or container with the necessary information, including the concentration and composition of the solution. Store the solution safely and securely, following any necessary storage precautions (if applicable).

By following these steps, a student can prepare 500.0 mL of a 0.08500 M solution of iron(III) using the given stock solution of 0.3180 M.