The conjugate acid of the base quinoline has $\mathrm{pK}_{\mathrm{a}}=4.88 .$ What fraction of molecules…

The conjugate acid of the base quinoline


has $\mathrm{pK}_{\mathrm{a}}=4.88 .$ What fraction of molecules protonated in a $0.010 \mathrm{M}$ aqueous solution of quinoline?
  1. $1.2 \times 10^{-3}$
  2. $2.8 \times 10^{-4}$
  3. $3.11 \times 10^{-4}$
  4. $1.87 \times 10^{-4}$

Solution

$\mathrm{pK}_{\mathrm{a}}=4.88 \Rightarrow \mathrm{pK}_{\mathrm{b}}=14-4.88=9.12$
or $\mathrm{K}_{\mathrm{b}}=10^{-9.12}=7.6 \times 10^{-10}$
Aqueous solution of quinoline establishes the following equilibrium.
$Q+H_{2} O ightleftharpoons Q H+O H^{-}$
$\begin{array}{lccc}\text { Quinoline } \\ \text { Initial } & 0.010 & 0 & 0 \\ \text { at equilibrium } 0.010-x & x & x\end{array}$
$\mathrm{K}_{\mathrm{b}}=\frac{\mathrm{x}^{2}}{0.010-\mathrm{x}} \mathrm{x}=-2.8 \times 10^{-6}$
$\mathrm{f}=\frac{\left[\mathrm{QH}^{+}ight]}{[\mathrm{Q}]_{\text {initial }}}=\frac{2.8 \times 10^{-6}}{0.010}=2.8 \times 10^{-4}$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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