Posted by Chris on Monday, March 5, 2012 at 8:50pm.
HONH2 I'll call RNH2
..........RNH2 + HOH ==> RNH3^+ + OH^-
initial...0.16M............0.......0
change......-x..............x......x
equil.....0.16-x............x......x
Kb = (RNH2)(OH^-)/RNH2
Substitute from the ICE chart above and solve for x which = OH^-, then convert to H^+ and pH.
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