A) \[0.6\times {{10}^{-4}}\]
B) \[6\times {{10}^{-4}}\]
C) \[6\times {{10}^{-5}}\]
D) \[3.6\times {{10}^{-4}}\]
Correct Answer: B
Solution :
Initial concentration = 0.006 M Þ \[{{K}_{a}}=6\times {{10}^{-5}}\] equilibrium reaction. \[\underset{C(1-a)\,\,\,}{\mathop{\underset{C}{\mathop{{{C}_{6}}{{H}_{5}}COOH}}\,}}\,\]⇌\[\underset{Ca}{\mathop{\underset{O}{\mathop{{{C}_{6}}{{H}_{5}}CO{{O}^{-}}}}\,}}\,+\underset{Ca\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{\underset{O\,\,\text{initial}}{\mathop{{{H}^{+}}\,\,\,\,\,\,\,}}\,}}\,\] \[{{K}_{a}}=C{{\alpha }^{2}}\] \[(\because 1-\alpha \approx 1)\]for weak electrolyte \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}\]; \[\alpha =\sqrt{\frac{6\times {{10}^{-5}}}{0.006}}\]; \[\alpha ={{10}^{-1}}\] \[\because \,[{{H}^{+}}]=C\alpha \]\[=0.006\times {{10}^{-1}}\]\[=6\times {{10}^{-4}}\]M.
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