First, we have to get Kb from this formula:
Pkb = - ㏒ Kb
by substitution by Pkb value = 5.79
5.79 = -㏒ Kb
∴Kb = 1.62x10^-6
From the ICE table
morphine ↔ C+ + OH-
initial 0.105 0 0
change -X +X +X
Equilibrium (0.105-X) X X
So Kb = [C+] [OH-]/[morphine]
by substitution
1.62x10^-6 = X^2 / (0.105-X)
X^2+ 1.6x 10^-6 X - 1.68x10^-7 / (X- 0.105) = zero by solving the equation for X
∴X =4 x 10 ^-4 or 0.0004
∴[OH-] = 0.0004
by substitution we can get POH
when POH = -㏒[OH-]
= -㏒0.0004
= 3.398
and when PH + POH = 14
∴PH = 14 - 3.398 = 10.6
