Question

Morphine is an effective pain killer but is also highly addictive. calculate the ph of a 0.105 m solution of morphine if its pkb = 5.79.

Answer 1

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