The pH 0.005 M codeine (${\mathrm{C}}_{18}{\mathrm{H}}_{21}{\mathrm{NO}}_3$) solution is 995. Its ${\mathrm{pK}}_{\mathrm{b}}$ value is

${\mathrm{C}}_{18}{\mathrm{H}}_{21}{\mathrm{NO}}_3 +{\mathrm{H}}_2\mathrm{O} \rightleftharpoons \text{ Codeine }{\mathrm{H}}^{+} +{\mathrm{OH}}^{-}$

$\mathrm{pH}=9.95 {\text{ P}}^{\mathrm{OH}} =14-9.95= 4.05$

${\mathrm{P}}^{\mathrm{OH}} = -\log [ {\mathrm{OH}}^{-}]$

$\log [ {\mathrm{OH}}^{-}]= -4.05= \overline{5}.95$

$\left[{\mathrm{OH}}^{-}\right]= \text{antilog }\overline{ 5}.95 = 8.913\times {10}^{-5}$

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[\text{ Codeine }{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[\text{Codeine }\right]}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{\left[\text{ Codeine }\right]}$

${\mathrm{K}}_{\mathrm{b}}=\frac{{\left(8.913\times {10}^{-5}\right)}^2}{0.005}=1.588\times {10}^{-6}$

${\mathrm{pK}}_{\mathrm{b}}=-\log \left[{\mathrm{K}}_{\mathrm{b}}\right] = -\log \left[1.588\times {10}^{-6}\right]$

${\mathrm{pK}}_{\mathrm{b}}=6 + (-0.2009)=5.7791\approx 5.80\approx 5.81$