Question

Nicotinic acid $(K_a={10}^{-5})$ is represented by the formula $HNiC$. Calculate percent dissociation in a solution which contains 0.1 mol of nicotinic acid per 2 litre of solution.

Answer 1

$HNiC\rightleftharpoons H^{+}+Ni{C}^{-}$

$C-C\alpha$ $C\alpha$ $C\alpha$

$K_a=\frac{(C\alpha {)}^2}{C(1-\alpha )}=\frac{C^2{\alpha }^2}{CC\mid \alpha \mid }=\frac{C{\alpha }^2}{1-\alpha }$

Since, it is a weak acid , $\alpha <<1$

$\therefore K_a=\frac{C{\alpha }^2}{1}$

$\therefore \alpha =\sqrt{\frac{Ka}{C}}$

Here, $C=\frac{0.1}{2}=0.05M,K_a={10}^{-5}$

$\alpha =\sqrt{\frac{{10}^{-5}}{0.05}}=\sqrt{2\times {10}^{-4}}=1.414\times {10}^{-2}$

$\therefore \%\alpha =1.414\%$