1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving Dilution Problem w/o Original Volume

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data

    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.
    2. Relevant equations

    molarity= moles/Liter

    3. 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.
     
  2. jcsd
  3. May 8, 2013 #2

    Borek

    User Avatar

    Staff: Mentor

    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.
     
  4. May 8, 2013 #3
    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*
     
  5. May 8, 2013 #4

    Borek

    User Avatar

    Staff: Mentor

    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.
     
  6. May 8, 2013 #5
    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.
     
  7. May 8, 2013 #6

    Borek

    User Avatar

    Staff: Mentor



    Check with C1V1=C2V2.
     
  8. May 8, 2013 #7
    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. :confused:
     
  9. May 8, 2013 #8

    Borek

    User Avatar

    Staff: Mentor

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

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

    OK

    Yes you do - you have already calculated much earlier how much 8M solution is needed (you even thought it should be a correct answer).
     
  10. May 8, 2013 #9
    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.
     
  11. May 8, 2013 #10

    Borek

    User Avatar

    Staff: Mentor

    Perfect.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solving Dilution Problem w/o Original Volume
Loading...