Question #7a067

1 Answer
il maestro
Mar 18, 2015

[H2NNH2]=2M [H2NNH3+]= 2.45 *10^-3M [OH-]= 2.45 *10^-3M

At equilibrium the concentration of N2H4 will be (2-x) M and the concentration of N2H5+ and OH- will be x M.
From the Kb we can write x^2 /(2-x) =3 *10^-6
Neglecting x in(2-x) since it is <<<<2, the equation simplifies in x^2 /2 =3 *10^-6 and resolving x=2.45 *10^-3

[OH-]=2.45 *10^-3 pOH=2.61 pH=11.39

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