question_answer

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) \[2.0\times {{10}^{-6}}\]       

Correct Answer: B

Solution :

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].