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Determining the molality of a solution

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data

    In a lab, 40.3 mL Mn2+ is mixed with 5 mL H3PO4. 0.4g of potassium periodate is added. The flask is heated and then diluted to 100mL. The final solution prepared is a permanganate solution. Calculate the final molarity of the solution.

    C Mn2+ = 0.06061 M (60.61 micrograms/ mL)
    C H3PO4 = 15 M
    V Mn2+ = 40.3 mL
    V H3PO4 = 5 mL
    M (Mn2+) = 54.94 g/mol
    V final = 100mL

    2. Relevant equations

    C1V1=C2V2

    2Mn2+ + 5IO4- + 3H2O ---> 2MnO4- + 5IO3- + 6H+

    3. The attempt at a solution

    (60.62 * 40.3 *10^-6) / (54.94 *0.100) = 4.45 *10^-4 M

    The 10^-6 was used as a conversion factor from micrograms to grams.

    I'm not sure if this is right, as it is not taking into account the H3PO4 added.
     
    Last edited: Dec 1, 2009
  2. jcsd
  3. Dec 1, 2009 #2

    Borek

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    Staff: Mentor

    What is the question?

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  4. Dec 1, 2009 #3
    Sorry! The question is to find the final molarity of the solution.
     
  5. Dec 1, 2009 #4

    Borek

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    Staff: Mentor

    Molarity of permanganate?

    How many moles of permanganate created per mole of manganese?

    Are you sure about it? M usually stands just for mol/L.

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    Last edited: Dec 1, 2009
  6. Dec 1, 2009 #5
    Yes, the molarity of permanganate. The concentration on Mn2+ was initially given in ppm (micrograms/mL) and i converted that value to mol/L...
     
  7. Dec 1, 2009 #6

    Borek

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    Staff: Mentor

    If you have converted 60.61 ppm to 0.06061 M you did it wrong.

    You have not answered my question.

    How many moles of manganese initially?

    How many moles of permanganate created per mole of manganese?

    What is the final volume?

    What is concentration definition?

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    methods
     
  8. Dec 1, 2009 #7
    I think i corrected my conversion : 60.61 ug/mL = 1.10 * 10^-3 M

    1. n = cv
    = 4.43 *10^-5
    2. 1:1 mole ratio
    3. final volume = 0.1 L
    4. concentration = 4.43*10^-4

    is the H3PO4 not involved in the reaction?
     
  9. Dec 2, 2009 #8

    Borek

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    Staff: Mentor

    I got 4.45x10-4M, that's most likely because of some rounding errors and/ordifferent molar masses used - but in general that's the same.

    Phosphoric acid... This is a limiting reagent problem. Think about amounts of substances used.

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