weak base BOH dissociate as
BOH + H2O B+(aq) + OH-(aq)
Kb = [B+] [OH-] / [BOH]
but [B+] = [OH-]
consider [B+] = [OH-] = X
then Kb = [X] [X] / [BOH]
Kb = X2 / [BOH]
X2 = Kb X [BOH]
substitute the value
X2 = (6.9 X 10-6) X (0.40) = 2.76 X 10-6
X = 0.00166 M
[B+] = [OH-] = X = 0.00166 M
[OH-] = 0.00166 M
pOH = -log [OH-] = -log (0.00166) = 2.78
pH = 14 - pH = 14 - 2.78 = 11.22
pH of 0.40 M weak base = 11.22
HW 14 © Alyson Smit 320 - WANG > Activities and Due Dates > HW 14...
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