Question

Codeine (C18H12NO3) is a weak base, (pKb= 5.79). A solution of 5*10^-2M solution has a pH of 10. Calculate the [C18H22NO3], [C18H21NO3] and [OH-] at equilibruim.

Answer 1

first find pOH from pH

pH + pOH = 14

pOH = 14 - pH    = 14 - 10 = 4

from the pOH we can calculate the [HO-]

[HO-] = 10^--4 = 0.0001 M

      C18H12NO3 (aq) + H2O (l) ----> C18H13NO3- (aq) + HO- (aq) construct ICE table

I       0.05 M                                         0                         0

C       -x                                               +x                       +x

E        0.05-x                                          +x                  +x

from the ICE table concentration of C18H13NO3- of concetration HO-

and we have alredy concentration of HO- = 0.001 M = [C18H13NO3-]

conceration of C18H12NO3 = 0.05 - x = 0.05 - 0.0001 = 0.0499 M