. Cocaine is a weak base, and the aqueous ionization can be indicated as:
C(aq) + H2O ≡ CH+(aq) + -
OH(aq)
In a 0.0040 M solution of cocaine, the ratio of -
OH(aq) ions to cocaine molecules is 1.0/120.
What is the value of Kb for cocaine?
.......................C(aq) + H2O ---> CH+(aq) + OH^-(aq)
I...........................................0..................0
C..............................................................
E................
Kb = (CH^+)(OH^-)/(C)
The problem tells you that for every 120 molecules of cocaine that the OH^-is 1.0 for a 0.004 M solution of cocaine. Therefore, as I see it,
(OH^-) = (CH^+) = 0.004 M x (1/120) = 3.33E-5 and (C) = 0.004 - 3.33E-5 = 0.004 - 0.000333 = 3.97E-3 Plug those values into Kb expression and solve. Post your work if you get stuck. I get something like 2.8E-7. I found on the web a value of 2.6E-7