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Chemistry-Finding concentrations of two compounds

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Given balanced chemical reaction:
    Ba(NO3)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2NaNO3 (aq)

    Suppose 200 mL of Ba(NO3)2 and 200 mL of Na2SO4 of unknown concentrations are mixed together and forms the reaction above. Then, there are 0.0120 M of NaNO3 produced, and 0.0004 M of excess sulfate ion. Find the initial concentration of each reactant.


    2. Relevant equations
    Molarity = number of moles/volume of solvent


    3. The attempt at a solution
    I led x be the concentration of Ba(NO3)2
    and y be the concentration of Na2SO4
    Then I used the law of conservation of mass to find the total mass of reactants and products, then wrote the equation below
    5.6x + 28.4y = 0
    But I am now not sure what to do next, do I need to find the volume of the solution produced or to do something else?
    Please give me some hints or concepts so that I can figure it out, I will be extremely appreciated...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 28, 2012 #2

    Borek

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    Staff: Mentor

    It can be solved in many ways, but the simplest approach is to look for things that have not changed. How many moles of NO3- in the final solution? How is this number related to the initial number of moles of Ba(NO3)2?
     
  4. Mar 28, 2012 #3
    I think I may get some ideas from what you asked
    I now try to write out the net ionic equation for the reaction below
    Ba2+ + 2NO3- + 2Na+ + SO42- → BaSO4 + 2Na+ + 2NO3-
    And I find out that NO3- and Na+ are spectator ions, that means their concentrations did not change right?
    But the problem is I don't know how to find the final concentrations of each ion from the given concentration of NaNO3 which is 0.0120M, should I divide it half and half, or is there some other ways??? Please advise..
     
  5. Mar 28, 2012 #4

    Borek

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    Staff: Mentor

    Concentration changes, but not because of teh reaction - you mix two solutions, so they are diluted.

    If concentration of NaNO3 is 0.0120M, concentrations of both Na+ and NO3- are also 0.0120M. This is not always the case, but here NaNO3 dissociates into 1+1 ions.
     
  6. Mar 29, 2012 #5
    So the final concentration of each ion is below
    Ba2+ = 0 M
    SO42- = 0.0004 M
    Na+ = 0.0120 M
    NO3- = 0.0120 M
    is it right?
    If they are right, then I will be able to find the initial concentration of Ba(NO3)2 by first finding the number of moles of NaNO3, and use the chemical equation to convert it to the number of moles of Ba(NO3)2, and finally divide it by 0.2 L to find out the initial concentration, did I think it right?
    But how about the initial concentration of another compound? I got confused by the excess 0.0004M sulfate acid
    Please advise, thank you.
     
  7. Mar 29, 2012 #6

    Borek

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    Staff: Mentor

    You are right about initial concentration of Ba(NO3)2.

    Part of the sulfate is in the precipitate (this part you can easily calculate from known initial concentration of barium), part of the sulfate is still in the solution. Sum of these was present in the initial solution of sodium sulfate.
     
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