chemistry 12

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A) Calculate [OH^-] in a o.12 M solution of hydrazine, H2NNH2(aq), having K of 3.0x10^-6.

B) what is the percent ionization of the H2NNH2(aq)

I can't get these for the life of mee!

  • chemistry 12 -

    If you will show your work I'll look for the error. I think OH^- is about 6E-4 and %ionization = about 0.50%

  • chemistry 12 -

    What i got was :

    a) K = (3.0x10^-6)(0.12)/(0.12)(0.12)= 0.6

    b) i got 1.90%

  • chemistry 12 -

    The error is you divided by 0.12 twice; once is all that is needed AND you didn't take the square root.
    N2H4 + HOH ==> N2H5^+ + OH^-

    Kb = (N2H5^+)(OH^-)/(N2H4)
    3.0E-6 = (x)(x)/(0.12)
    x^2 = sqrt(3.0E-6*0.12)
    I obtained 6.0E-4 for OH^-

    Then %ion = [(OH^-)/(0.12)]*100 = ?
    I obtained 0.50%

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