Question #7a067

1 Answer
il maestro
Mar 18, 2015

[H2NNH2]=2M [H2NNH3+]= 2.45 *10^-32.45⋅10−3M [OH-]= 2.45 *10^-32.45⋅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^2x2 /(2-x)(2−x) =3 *10^-63⋅10−6
Neglecting xx in(2-x)(2−x) since it is <<<<2⟨⟨2, the equation simplifies in x^2x2 /22 =3 *10^-63⋅10−6 and resolving xx=2.45 *10^-32.45⋅10−3

[OH-]=2.45 *10^-32.45⋅10−3 pOH=2.612.61 pH=11.3911.39

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