Posted by Nancy on Wednesday, April 28, 2010 at 10:50am.
BOH ==> B^+ + OH^-
Kb = (B^+)(OH^-)/(BOH)
pH = 10.2; therefore, pOH = 3.8
pOH = -log(OH^-). Solve for OH^- and substitute into Kb expression. For (BOH) substitute 0.045-(OH^-).
Post your work if you get stuck.
For pOH=-log(OH-) I solved to get OH=1.58e-4
then I did kb=(4.50e-2)(1.58e-4)/.045-1.58e-4 to get 1.59e-4 but that's not the right answer
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