• # question_answer 56) A 0.010 M solution of maleic acid, a monoprotic organic acid is 14 per cent ionized. What is the${{K}_{a}}$for maleic acid? A) $2.3\times {{10}^{-3}}$                B) $2.3\times {{10}^{-4}}$C) $2.0\times {{10}^{-4}}$                D) $2.0\times {{10}^{-6}}$

Degree of dissociation, $\alpha =\frac{14}{100}=0.14.$ Concentration of the maleic acid, C = 0.010 M. The equilibrium reaction is Hence, the correct option is .  $t=0$ $C$ 0 0 ${{t}_{eq}}$ $C(1-\alpha )$ $C\alpha$ $C\alpha$
${{K}_{a}}=\frac{{{C}^{2}}{{\alpha }^{2}}}{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}$ ${{K}_{a}}=\frac{0.010\times 0.14\times 0.14}{0.86}$ ${{K}_{a}}=2.3\times {{10}^{-4}}.$ Hence, the correct option is [b].