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Equilibrium — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.
✓

Answer

c = 0.005
pH = 9.95
pOH = 4.05
pH = – log (4.105)
$4.05=-\log[\text{OH}^-]$
$[\text{OH}^-]=8.91\times10^{-5}$
$\text{c}\alpha=8.91\times10^{-5}$
$\alpha=\frac{8.91\times10^{-5}}{5\times10^{-3}}=1.782\times10^{-2}$
$\text{Thus, K}_\text{b}=\text{c}\alpha^2$
$=0.005\times(1.782)^2\times10^{-4}$
$=0.005\times3.1755\times10^{-4}$
$=0.0158\times10^{-4}$
$\text{K}_\text{b}=1.58\times10^{-6}$
$\text{Pk}_\text{b}=-\log\text{K}_\text{b}$
$=-\log(1.58\times10^{-6})$
$=5.80$

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