What is the value of Kb for coccaine?

Question

The much-abused drug cocaine is an alkaloid. Alkaloids are noted for their bitter taste, an indication of their basic properties. Cocaine, C12 H21 O4 N, (MM: 303.491) is soluble in water to the extent of 0.17 g/100 mL solution, and a saturated solution has a pH=10.08.

C17H21O4N + H2O <-> C17H21O4NH+ + OH- Kb = ?

Answer 1

0.17 g/100 mL

0.17g/303.491=0.0005601 mol

100mL/1000=0.100 L

[C17H21O4N]=0.0005601 mol / 0.100 L = 0.005601 M

C17H21O4N + H2O <-> C17H21O4NH+ + OH-

I 0.005601 0 0

C -x +x +x

E 0.005601-x x x

Since pH at equilibrium = 10.08,

pH+pOH=14

14-10.08=3.92

pOH=3.92

[OH-]=10^-pOH

[OH-]=10^-3.92=0.0001202

Plug this in for x in the E row to solve for equilibrium concentrations

[C17H21O4N]=0.005601-x

=0.005601-0.0001202

=0.0054808 M

[OH-]=[C17H21O4NH⁺]=x=0.0001202

Kb=([OH-][C17H21O4NH⁺]) / [C17H21O4N]

Kb=(0.0001202×0.0001202)/(0.0054808)

Kb=0.00000263611=2.64×10¯⁶