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.
Calculate the [H+], [OH ‾ ], and the pH of 0.16 M solutions of:
hydroxylamine (HONH2, Kb = 1.1 10-8)
1 answer