Posted by Chris on Wednesday, December 5, 2012 at 6:49pm.
Let's call diethylamine simply BNH.
.......BNH + HOH ==> BNH2^+ + OH^-
I.....0.600...........0........0
C........-x............x........x
E......0.600-x........x.........x
%ionization = 4.65%; therefore,
x = (BNH^+) = (OH^-) = 0.600 x 0.0465
pOH = -log(OH^-) and yo have OH^-
Then pH + pOH = pKw = 14. You have pOH and 14, solve for pH.
Then substitute the values in the E line into Kb expression and solve for Kb.
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