Question #7a067

1 Answer
il maestro
Mar 18, 2015

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

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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