Posted by joane on Friday, November 30, 2012 at 1:56pm.
pKb = -log Kb
-8.75 = log Kb
Solve for Kb.
........BN + H2O ==> BNH^+ + OH^-
I......0.190..........0.......0
C.......-x............x........x
E......0.190-x........x........x
Substitute the E line into the Kb expression and solve for x = OH^-, then convert to pH.
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