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The 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 \(\mathrm{pK}_{\mathrm{b}}\)
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\(\mathrm{Cod}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CodH}^{+}+\mathrm{OH}^{-}\)
\(\mathrm{pH}=9.95 \therefore \mathrm{pOH}=14-9.95=4.05\),
\([\mathrm{pH}+\mathrm{pOH}=14]\)
i.e., \(-\log \left[\mathrm{OH}^{-}\right]=4.05\)
or \(\log \left[\mathrm{OH}^{-}\right]=-4.05=-5.95\)
or \(\left[\mathrm{OH}^{-}\right]=8.913 \times 10^{-5}\)
\(\mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{CodH}^{+}\right][\mathrm{OH}]}{[\mathrm{Cod}]}=\frac{\left[\mathrm{OH}^{-}\right]^2}{[\mathrm{Cod}]}=\frac{\left(8.91 \times 10^{-5}\right)^2}{5 \times 10^{-3}}\)
[at equilibrium \(\left[\mathrm{Cod} \mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]\)
\(\mathrm{pK}_{\mathrm{b}}=-\log \left(1.588 \times 10^{-6}\right)\)
\(=-\log 1.59+6 \log 10\)
\(=6-\log 1.59=5.80\)
\(\mathrm{pH}=9.95 \therefore \mathrm{pOH}=14-9.95=4.05\),
\([\mathrm{pH}+\mathrm{pOH}=14]\)
i.e., \(-\log \left[\mathrm{OH}^{-}\right]=4.05\)
or \(\log \left[\mathrm{OH}^{-}\right]=-4.05=-5.95\)
or \(\left[\mathrm{OH}^{-}\right]=8.913 \times 10^{-5}\)
\(\mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{CodH}^{+}\right][\mathrm{OH}]}{[\mathrm{Cod}]}=\frac{\left[\mathrm{OH}^{-}\right]^2}{[\mathrm{Cod}]}=\frac{\left(8.91 \times 10^{-5}\right)^2}{5 \times 10^{-3}}\)
[at equilibrium \(\left[\mathrm{Cod} \mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]\)
\(\mathrm{pK}_{\mathrm{b}}=-\log \left(1.588 \times 10^{-6}\right)\)
\(=-\log 1.59+6 \log 10\)
\(=6-\log 1.59=5.80\)
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