question_answer

If 0.1 M of a weak acid is taken, and its percentage of degree of ionization is 1.34%, then its ionization constant will be :

A) \[0.8\times {{10}^{-5}}\]         

B)        \[1.79\times {{10}^{-5}}\]

C) \[0.182\times {{10}^{-5}}\]      

D)        none of the above

Correct Answer: C

Solution :

[c] Percentage of degree of ionization = 1.34% \[\therefore \] Degree of ionization \[(\alpha )=0.0134\] \[K=\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}\]     \[=\frac{0.1\times 0.0134\times 0.0134\times 0.1}{0.1\times (1-0.0134)}\] \[=0.182\times {{10}^{-5}}\]