Question

The $pH$ of 0.005 M codeine ($C_{18}{H}_{21}{NO}_3$) solution is 9.95. Calculate its ionization constant and ${pK}_b$.

Answer 1

$pH=9.95$,

$pOH=14-pH=14-9.95=4.05$

$\left[O{H}^{-}\right]={10}^{-pOH}={10}^{-4.05}=8.913\times {10}^{-5}$

$Codeine+H_{2}O\rightleftharpoons Codeine{H}^{+}+O{H}^{-}$

The ionization constant, $K_b=\frac{\left[Codeine{H}^{+}\right]\left[O{H}^{-}\right]}{\left[codeine\right]}=\frac{(8.913\times {10}^{-5})\times (8.913\times {10}^{-5})}{5\times {10}^{-3}}=1.588\times {10}^{-6}$.

$p{K}_b=-log(1.588\times {10}^{-6})=5.8$