lets B is the base
construct ICE table
    B (aq) + H2O (l) <-----> BH (aq)
+ OH- (aq)
IÂ Â Â Â
0.39Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â
-x                                    Â
+x             Â
+x
EÂ Â
0.39-x                              Â
+x              Â
+x
Kb = [BH] [OH-] / [B]
3.8 x 10^-6 = [x] [x] / [0.39-x]
x2 + x 3.8 * 10^-6 - 1.482 * 10^-6 = 0
solve the quadratic equation
x = 0.00121 M = [OH-]
pOH = -log[OH-] = -log[0.00121] = 2.91
pH + pOH = 14
pH = 14-pOH = 14 - 2.91 = 11.01
