Dilution and moles

Discussion in 'Chemistry' started by chmdummy1982, Mar 1, 2009.

  1. chmdummy1982 Registered Member

    Messages:
    1
    I am having trouble with a couple of chemistry problems. One I was able to answer half of, and the other I don't even know how to start.


    Question 1

    Use the dilution relationship (Mi x Vi = Mf x Vf) to calculate the volume of 0.500 M NaOH needed to prepare 500 mL of 0.250 M NaOH.

    ? mL x .500 M NaOH = (500mL x.250M)/.500 = 250 mL


    Now here is where I am lost with this problem:

    What volume of water would you add to the 0.500 M NaOH volume calculated aove to actually make 500 mL of 0.250 M NaOH?



    Question 2
    49.22 mL of a 2.01 M NaOH solution reacts completely with 40.28 mL of HCl solution according to the balanced chemical reaction shown below:

    HCl (aq) + NaOH(ap) -> NaCl(aq) + H2O(i)

    How many moles of HCl reacted?

    ^ I don't even know where to began with this one, here is my best attempt.

    ? mol HCl = 49.22 mL NaOH x (2.01mNaOH/1 mL NaOH) x (40.28 mL Hcl/ 1 m NaOH) x (1 mol Hcl/1 mL HCl)
     
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  3. Dark520 Rebuilt Registered Senior Member

    Messages:
    403
    Question 1:
    Your final solution needs to be 500mL and .250M, your initial is of unknown volume and .500M. So take that info and plug it in:
    ?mL x .500M = 500mL x .250M --> ? = 250mL

    So, you have 250mL of .500M NaOH (which is .125 moles), and your final solution needs to be 500mL, so just add 250mL of water and you're good to go.

    Question 2:
    We know that molarity = moles / liter, and you have the molarity and liters of NaOH. Because this reaction is 1 for 1 (1 mol NaOH neutralizes 1 mol HCl), the mols of NaOH and HCl have to be equal. That means that if we find the moles NaOH, we've also found the moles of HCl. In order to do that, we plug what we know into a rearranged form of the equation I stated above: moles = molarity * liters --> moles = 2.01M * .04922L = .0989 mols. Mols NaOH = mols HCl, so .0989 moles HCl were used.

    Even though it's not asked, it might be a good idea to find the molarity of that HCl solution. Plugging the info we have back into the molarity formula: molarity = .0989 moles / .04028L = 2.46M HCl.

    Also, keep in mind that we could only equate the moles NaOH to moles HCl because they react 1 to 1. If we had used H2SO4 (sulfuric acid), then they would react 2 to 1, because H2SO4 can release 2 H+ ions. In that case, 2 mols of NaOH react to neutralize 1 mole of H2SO4. So, you''d have to divide the moles NaOH by two in order to find the moles H2SO4.
     
    Last edited: Mar 1, 2009
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  5. Trippy ALEA IACTA EST Staff Member

    Messages:
    10,890
    You've got half the amount of final solution that you need, but it's twice as concentrated as you need it to be. What do you think the other half of the solution might be? In other words - at the moment, if you were to have done it this far in a lab - IE: extracted the 250 ml of Sodium Hydroxide solution, but not yet added water to it, what you would actually have is 250 ml of 0.500 M NaOH.



    From the balanced equation, we have a 1 to 1 relationship between the number of moles of Hydrochloric acid used.
    So you need to use the equation n=c*v (or, I think in your case you're using n=M*v) to find out how many moles of sodium hydroxide were reacted.
    Then for every mole of sodium hydroxide reacted, you have one mole of hydrochloric acid reacted.
    You only need the volume of Hydrochloric acid solution if your being asked to calculate the concentration of the hydrochloric acid solution (which is usually one of the next questions in this sort of question set).

    Oh, and one last thing.

    When doing your calculations to find the number of moles, DON'T FORGET TO CONVERT YOUR VOLUMES TO LITRES!
     
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