Answer:
Kb = 1.6*10^-6
Explanation:
The given reaction is:
[tex]N2H4(aq)+H2O(l)\rightarrow N2H5+(aq)+OH-(aq)[/tex]
The ionization constant of the base Kb is given as:
[tex]Kb = \frac{[N2H5+][OH-]}{[N2H4]}------(1)[/tex]
The pH = 10.66
therefore, pOH = 14-pH = 14-10.66 =3.34
[tex][OH-] = 10^{-pOH} =10^{-3.34} =4.57*10^{-4} M[/tex]
[N2H5+] = [OH-] = 4.57*10^-4M
[N2H4] = 0.133 M
Based on eq(1)
[tex]Kb = \frac{[4.57*10^{-4}]^{2}}{[0.133]}=1.6*10^{-6}[/tex]
