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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 51 of 73
Marks: +1, -0
The pH of 0.005 M codeine (C18H21NO3)(\mathrm{C}_{18}\mathrm{H}_{21}\mathrm{NO}_3) solution is 9.95. Calculate its ionization constant and pKb\mathrm{p}K_b.
Solution:  
[H+][\mathrm{H}^{+}] = antilog (- 9.95) = 1.12 × 10−1010^{-10}
∴ [OH−][\mathrm{OH}^{-}] = 1.0×10−141.12×10−10\frac{1.0 \times 10^{-14}}{1.12 \times 10^{-10}} = 8.92 × 10−510^{-5} M
But Kb\mathrm{K}_b = [M+][OH−][MOH]\frac{[\mathrm{M}^{+}][\mathrm{OH}^{-}]}{[\mathrm{MOH}]} = (8.92×10−5)20.005\frac{(8.92 \times 10^{-5})^2}{0.005} = 1.59 × 10−610^{-6}
pKb\mathrm{p}K_b = - log Kb\mathrm{K}_b = - log (1.59 × 10−610^{-6}) = 5.80
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