Question

Nicotinic acid $(K_a=1.4\times {10}^{-5})$ is represented by the formula $HNiC$. Calculate its percent dissociation in a solution which contains $0.10$ mole of nicotinic acid per $2$ litre of solution.

Answer 1

$HNiC\rightleftharpoons H^{+}+{NiC}^{-}$
Equi. conc. $(0.05-x)$ $x$ $x$

As $x$ is very small, $(0.05-x)$ can be taken as $0.05$
$K_a=\frac{\left[H^{+}\right]\left[{NiC}^{-}\right]}{\left[HNiC\right]}=\frac{x\times x}{0.05}$
or $x^2=\left(0.5\right)\times (1.4\times {10}^{-5})$
$x=0.83\times {10}^{-3}mol{L}^{-1}$
% dissociation$=\frac{0.83\times {10}^{-3}}{0.05}\times 100=1.66$