For the reaction of hydrazine (N2H4) in water, Kb is 3.0 10-6.
H2NNH2(aq) + H2O(l)--> H2NNH3+(aq) + OH-(aq)
Calculate the concentrations of all species and the pH of a 1.6 M solution of hydrazine in water.
[OH-] =
[H2NNH3+] =
[H2NNH2] =
[H+] =
pH =

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  1. .....H2NNH2 + HOH ==> H2NNH3^+ + OH^-

    Kb = 3.0E-6 = (H2NNH2^+)(OH^-)/(H2NNH2)
    Substitute the E line into the Kb expression and solve for x.
    That will give you H2NNH3^+, H2NNH2, OH^-. From that you can calculate pOH = -log(OH^^-), then get pH from
    pH + pOH = pKw = 14

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