1. #1

    Quick Chemistry Help

    Just asking real quick to see if anybody can help me on this...

    A 15.00-mL aqueous sample was analyzed for its nitrite content by reaction with 40.13 mL of 0.04013 N acidified potassium permanganate solution, liberating nitrate ion. The excess permanganate was determine by reaction with 0.04013 N iron(II) solution, requiring 2.59 mL to reach equivalence. Determine the g/L concentration of nitrite in the original sample.

    Anything you guys can offer would be appreciated

  2. #2
    I would suggest reviewing your notes and textbook and then do your own homework.

    Let's all ride the Gish gallop.

  3. #3
    N is not M for molarity. N is for Normality, a right pain in the ass of a measurement.

  4. #4
    Quote Originally Posted by Sociopathic View Post
    N is not M for molarity. N is for Normality, a right pain in the ass of a measurement.
    It is,but in this case the assumption of N = mg/L is most likely adequate.

    Then calculate the mass of Fe and permanganate. C1V1=C2V2

    - - - Updated - - -

    Quote Originally Posted by Aeriel View Post
    It is,but in this case the assumption of N = mg/L is most likely adequate.

    Then calculate the mass of Fe and permanganate. C1V1=C2V2
    Side note: Nitrate/Nitrite analysis is one of the analyses that regularly run, but this is not the method I use.
    Last edited by Aeriel; 2015-06-15 at 02:10 AM.

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