Solving Dilution Problem w/o Original Volume

[JeweliaHeart]

Homework Statement

What volume of water must be added to a sample of 8.00 M Nitric Acid in order to produce 500. mL of 1.25 M Nitric Acid?

(A) 422 mL
(B) 499 mL
(C) 625 mL
(D) 0.625 L
(E) None of The Above

The correct answer is A in bold.

Homework Equations

molarity= moles/Liter

The Attempt at a Solution

This is problem 1 from:
http://www.adriandingleschemistrypages.com/apquiz04D.html

.5 L(500. mL ) x 1.25 moles HNO3/Liter solution=0.625 moles HNO3

(0.625/.5)=8/(1+x) The reason I left it as 1+x is b/c I assumed the volume of the solution is already 1 Liter and therefore x represents the amount of water that needs to be added in order to reduce the concentration to (.625/.5)

The answer I am left with is x=5.4, which is not one of the choices. I think that maybe I should not have assumed a volume of 1 L, but that the problem should say what the original volume is...But maybe not. I'm not sure what I did wrong.

 

[Borek]

Why do you assume 1L if you are asked to produce 0.5L?

Calculate how much acid you need and what volume of 8M solution will contain that amount.

 

[JeweliaHeart]

The reason I assumed a volume of 1 Liter is b/c it seems like a dilution problem. It asks what volume of water must be added to a sample of solution, implying that there already is a certain volume of solution and we must dilute it in order to get to a certain concentration
(1.25 M). The problem with my theory though is that there would be no reason for the problem to mention the "0.500 mL of 1.25 M Nitric Acid" part.

Initially, I tried what you said and then arrived at the dilution theory b/c it didn't work out.

Here's what I did:

0.5 L x 1.25 moles/L =0.625 moles

8 moles/Liter x (V) =0.625 moles (V stands for volume of solution)

V= 0.078125 L (not an answer choice)


I must've messed up in somewhere or another. *scratches head*

 

[Borek]

V= 0.078125 L (not an answer choice)

Because it is not yet the final answer. But you are close.

This is just a necessary volume of 8M nitric acid. Now add enough water to make it 0.5L.

 

[JeweliaHeart]

Okay so

0.5-0.078125= 0.422

Yes! Thank You...

But does this really give 0.5 L of a 1.25 M Nitric Acid solution like the problem asks?

I feel like I played with numbers without really understanding what I meant to accomplish with them.

 

[Borek]

But does this really give 0.5 L of a 1.25 M Nitric Acid solution like the problem asks?

Check with C₁V₁=C₂V₂.

 

[JeweliaHeart]

C1=1.25 C2=(8/0.5)
V1=0.5 V2= ?


(1.25)(0.5)=(8/0.5)V2

I don't know Volume 2. All we found out in the problem was the new concentration once the solution was diluted. I don't know what the new volume is. This problem is awkward.

 

[Borek]

C1=1.25

I would reverse the indices, so that it would be my C₂, but OK, I will follow this convention.

C2=(8/0.5)

No, it should be original 8M. Dividing concentration by volume doesn't yield anything reasonable.

V1=0.5

OK

I don't know Volume 2.

Yes you do - you have already calculated much earlier how much 8M solution is needed (you even thought it should be a correct answer).

 

[JeweliaHeart]

Ok, I reviewed the question. I think I understand better now. We are trying to find what volume of solution must be added to solution that has a concentration of 8 moles/Liter in order that the new solution will be made up of 0.625 moles solute and 0.5 Liters solution.

So, I should've written:

C1=8 moles/L

V1=0.078125 L

C2=1.25 moles/L

V2=0.5


8(0.0.78125)=1.25(0.5) ---------> both of which equal 0.625

So the C1V1=C2V2 tells us that both solutions have 0.625 moles at their initial volumes and therefore when we added the 0.422 L we raise the volume to 0.5 L and create a concentration equal to the 1.25 M solution.

Sorry if that is too wordy. I just had to write out to make sure what I was thinking in my head made sense at all. I think I understand better now. Thank you so much. You have been a big help.

 

[Borek]

Perfect.