Posted by Lori on Saturday, February 14, 2009 at 9:43pm.
You do these the same way except with Ka you are dealing with the ionization of an acid and with Kb you are dealing with the base and the reaction with water.
Kb = (H2NNH3^+)(OH^-)/(H2NNH2)
Do an ICE chart and solve for (OH^-).
Post your work if you get confused along the way. My answer, without using the quadratic equation, is about 2.6 for the pOH.
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