A) \[1.8\times {{10}^{-3}}M\]
B) \[1.8\times {{10}^{-5}}M\]
C) \[1.3\times {{10}^{-3}}M\]
D) \[1.34\times {{10}^{-3}}M\]
Correct Answer: D
Solution :
\[C{{H}_{3}}COOH\]ionise as follows : \[\underset{C(1-\alpha )}{\mathop{C{{H}_{3}}COOH}}\,\underset{C\alpha }{\mathop{C{{H}_{3}}CO{{O}^{-}}}}\,+\underset{\begin{smallmatrix} C\alpha \\ Final\,conc. \end{smallmatrix}}{\mathop{{{H}_{3}}{{O}^{+}}}}\,\] \[[{{H}_{3}}{{O}^{+}}]=C\alpha \] and \[\alpha =\sqrt{{{K}_{a}}/C}\] (for any weak electrolyte) or \[[{{H}_{3}}{{O}^{+}}]=\sqrt{{{K}_{a}}.C}\] \[=\sqrt{1.8\times {{10}^{-5}}\times .1}=1.34\times {{10}^{-3}}M\]
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