A chemist is given an unknown base. She finds that a 1.0461 M aqueous solution of this base has a pH of 13.065 at 25oC. What is the value of Kb for this unknown base ?.
pH = 13.065 = pOH = 0.935
pOH = -log(OH^-) and (OH^-) = 0.116
........BOH ==> B^+ + OH^-
initial.1.0461...0.......0
change...-x.....x.......x
equil.1.0461-x...x.....x
Kb = (B^+)(OH^-)/(BOH)
Substitute the ICE chart data into Kb expression and solve for Kb.
(B) = 0.1161 = (OH^-) and (BOH) = 1.0461-0.1161