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A problem in Redox Titrations (Kjeldhal's Method)

  1. Jun 1, 2017 #1
    1. The problem statement, all variables and given/known data:

    1.08g of ##NH_4 Cl## is boiled with 62ml ##\frac {N}{2}## ##NaOH## to expel ##NH_3## completely. 44ml of ##\frac {N}{4}## ##HCl## is required to neutralise the excess alkali. Calculate the percentage of ##NH_3 ## in ##NH_4 Cl##.

    Answer given: 31.5%

    2. Relevant equations:

    3. The attempt at a solution:

    I have found out the number of equivalents of ammonia liberated, which is the same as the number of equivalents of ammonium chloride used, and that of the number of equivalents of ##NaOH## that reacted, keeping in mind to subtract the excess number of equivalents of the alkali.

    But, how do I find the weight of ammonia used? In the reaction of ##NH_4 Cl## with ##NaOH##, the oxidation number of Nitrogen doesn't change. So, what is the valency factor or 'n' factor for ammonia to find the equivalent weight? Should I just use Nitrogen's imaginary oxidation number in Ammonia to find the equivalent weight?

    Or should I follow a completely different approach?
     
  2. jcsd
  3. Jun 1, 2017 #2

    Borek

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    Staff: Mentor

    Hard to say what it is about. A bit nonsensical question (not your fault). Very convoluted way of finding out that

    [tex]\frac {M_{NH_3}} {M_{NH_4Cl}} \times 100\% = 31.5 \%[/tex]
     
  4. Jun 1, 2017 #3
    What does the M mean?
     
  5. Jun 1, 2017 #4

    Borek

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    Staff: Mentor

    Molar mass.
     
  6. Jun 1, 2017 #5
    OK, so how should I find Molar Mass from number of equivalents?
     
  7. Jun 1, 2017 #6

    Borek

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    Staff: Mentor

    You don't have to, which is why I think this question doesn't make much sense. Most of the information given is completely useless.
     
  8. Jun 1, 2017 #7
    So, I can conclude that the question is nonsensical, and I mark it solved. Anyways, thank you for your help. :smile:
     
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