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

The congugate 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

$\mathbf{p K}_{\mathrm{a}}=4.88 \Rightarrow \mathrm{pK}_{\mathrm{b}}=14-4.88=9.12$
or $\mathrm{K}_{\mathrm{b}}=10^{-9.12}=7.6 \times 10^{-10}$
Aqueous solution of quinoline establish following equilibrium.
$\mathbf{Q}+\mathbf{H}_{2} \mathrm{O} ightleftharpoons \mathbf{Q}^{+}{ightleftharpoons} \mathbf{Q}+\mathbf{O H}$
(Quinoline)
$\begin{array}{ll}\text { Initial } & \mathbf{0 . 0} 10\end{array}$
at equilibrium $0.010-x$ $x \quad x$
$K_{b}=\frac{x^{2}}{0.010-x} x=-2.8 \times 10^{-6}$
$\mathrm{f}=\frac{\left[\mathrm{QH}^{+}ight]}{[\mathrm{Q}]_{\text {intial }}}=\frac{2.8 \times 10^{-6}}{0.010}=2.8 \times 10^{-4}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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