Posted by rodrigue on Friday, February 19, 2010 at 7:46pm.
BOH ==> B^+ + OH^-
Kb = (B^+)(OH^-)/(BOH)
You know pH, convert to pOH, then to (OH^-) and substitute into Ka expression. For (BOH), substitute 1.2 - (OH^-) and solve for Kb.
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