Dilution of a Solution Strengths

[frgeorgeh]

I want to mix 4 strengths of solutions. I thought to save time I would mix 1 a 0mg strength and a second at 18mg strength. How do I use those two solutions to make a 6mg, 12 mg strength solution?

The ratio of (OV)*(OS)=(NV)*(NS) where OV=Old Volume, OS=Old Strength,, NV=New Volume, NS=New Strength, does not work since I know all the results. For example, if I wanted to make the following:

I have: 1 gal (3,785ml) of 0mg and 1gal of 18mg

I want to make: 30ml bottles at 6mg and 12mg.

I hope this makes sense.

Thanks

 

[frgeorgeh]

I want to mix 4 strengths of solutions. I thought to save time I would mix 1 a 0mg strength and a second at 18mg strength. How do I use those two solutions to make a 6mg, 12 mg strength solution?

The ratio of (OV)*(OS)=(NV)*(NS) where OV=Old Volume, OS=Old Strength,, NV=New Volume, NS=New Strength, does not work since I know all the results. For example, if I wanted to make the following:

I have: 1 gal (3,785ml) of 0mg and 1gal of 18mg

I want to make: 30ml bottles at 6mg and 12mg.

I hope this makes sense.

Thanks

__________________________________

OK, I think I am over thinking this. It seems to me, to get a 6mg strength, I add 2/3 of the 0MG and 1/3 of the 18mg. To get 12mg, I add 1/3 0mg and 2/3 of 18mg. What do u think?

 

[MarkFL]

I am assuming the solution strengths are in milligrams per milliliters. So, if you have a solution with a strength of $18$, to get a solution of volume $V$ with a strength of $18k$, where:

\(\displaystyle 0\le k\le1\)

You would need to use \(\displaystyle kV\) of the solution of strength $18$, with the rest consisting of the strength $0$ solution.

For example, since 12 is 2/3 of 18, then you would put 2/3 of 30 mL which is 20 mL with 10 mL of strength 0 to get 30 mL of strength 12.

So yes, your reasoning is correct. :D

 

[frgeorgeh]

I am assuming the solution strengths are in milligrams per milliliters. So, if you have a solution with a strength of $18$, to get a solution of volume $V$ with a strength of $18k$, where:

Thanks for the reply! I am an Engineer but always seem to over complicate my math.

 

[CellCoree]

for your question! Dilution of solutions is a common practice in the scientific world, and it can be a bit tricky to understand at first. However, with some basic knowledge of chemistry and math, you can easily calculate the volumes needed to create your desired solution strengths.

First, let's define some terms for clarity:

- Old Solution: This refers to the 0mg and 18mg solutions that you have on hand.
- New Solution: This refers to the 6mg and 12mg solutions that you want to create.
- Volume (V): This is the amount of solution you have or want to create, measured in liters (L) or milliliters (mL).
- Strength (S): This is the concentration of the solution, typically measured in milligrams per milliliter (mg/mL).

Now, to calculate the volumes needed for your desired solutions, we can use the following formula:

(V1)(S1) = (V2)(S2)

Where:

- V1 = Volume of the old solution (in mL)
- S1 = Strength of the old solution (in mg/mL)
- V2 = Volume of the new solution (in mL)
- S2 = Strength of the new solution (in mg/mL)

So, let's plug in the values from your example:

For the 6mg solution:

(3785 mL)(0 mg/mL) = (V2)(6 mg/mL)

V2 = (3785 mL)(0 mg/mL) / (6 mg/mL)

V2 = 0 mL

This means that in order to create a 6mg solution, you will need to mix 0 mL of the 0mg solution with 3785 mL of the 18mg solution.

For the 12mg solution:

(3785 mL)(0 mg/mL) = (V2)(12 mg/mL)

V2 = (3785 mL)(0 mg/mL) / (12 mg/mL)

V2 = 0 mL

Similarly, you will need to mix 0 mL of the 0mg solution with 3785 mL of the 18mg solution to create a 12mg solution.

I understand that this may seem counterintuitive, as you may expect to use some of the 0mg solution to create the 6mg and 12mg solutions. However, since the 0mg solution has no strength, it essentially has no effect on the overall strength of the solution