Question:
The $\mathrm{pH}$ of $0.005 \mathrm{M}$ codeine $\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)$ solution is $9.95$. Calculate its ionization constant and $\mathrm{pK}_{\mathrm{b}}$.
Solution:
c = 0.005
pH = 9.95
pOH = 4.05
pH = – log (4.105)
$4.05=-\log \left[\mathrm{OH}^{-}\right]$
$\left[\mathrm{OH}^{-}\right]=8.91 \times 10^{-5}$
$c \alpha=8.91 \times 10^{-5}$
$\alpha=\frac{8.91 \times 10^{-5}}{5 \times 10^{-3}}=1.782 \times 10^{-2}$
Thus, $K_{b}=c \alpha^{2}$
$=0.005 \times(1.782)^{2} \times 10^{-4}$
$=0.005 \times 3.1755 \times 10^{-4}$
$=0.0158 \times 10^{-4}$
$K_{b}=1.58 \times 10^{-6}$
$P k_{b}=-\log K_{b}$
$=-\log \left(1.58 \times 10^{-6}\right)$
$=5.80$