Answer
$C_{18}H_{21}O_3NH^+(aq) + H_2O(l)$ $C_{18}H_{21}O_3N(aq) + H_3O^+(aq)$
$pKb = 7.95$
Work Step by Step
1. Write a reaction where protonated codeine acts like an acid:
$C_{18}H_{21}O_3NH^+(aq) + H_2O(l)$ $C_{18}H_{21}O_3N(aq) + H_3O^+(aq)$
** It needs to be donating one proton.
2. As we can see, codeine and protonated codeine are a conjugate acid-base pair, therefore:
$pKa + pKb = pKw = 14$
$6.05 + pKb = 14$
$pKb = 7.95$
