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NEET Sample Paper NEET Sample Test Paper-37

  • question_answer
    Ionisation constant of \[C{{H}_{3}}COOH\]is \[1.7\times {{10}^{-5}}\]and concentration of \[{{H}^{+}}\]ions is \[3.4\times {{10}^{-4}}.\] Then, find out initial concentration of \[C{{H}_{3}}COOH\]molecules.

    A) \[3.4\times {{10}^{-4}}\]

    B) \[3.4\times {{10}^{-1}}\]

    C) \[6.8\times {{10}^{-4}}\]

    D) \[6.8\times {{10}^{-3}}\]

    Correct Answer: D

    Solution :

     \[C{{H}_{3}}COOH\rightleftharpoons C{{H}_{4}}COO+{{H}^{+}}\] \[{{K}_{a}}=\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}\] Given that \[[C{{H}_{3}}CO{{O}^{-}}]=[{{H}^{+}}]=3.4\times {{10}^{-4}}\,M\] \[{{K}_{a}}\]for \[C{{H}_{3}}COOH=1.7\times {{10}^{-5}}\] \[C{{H}_{3}}COOH\]is weak acid, so in it \[[C{{H}_{3}}COOH]\] is equal to initial concentration. Hence, \[1.7\times {{10}^{-5}}=\frac{(3.4\times {{10}^{-4}})(3.4\times {{10}^{-4}})}{[C{{H}_{3}}COOH]}\] \[[C{{H}_{3}}COOH]=\frac{3.4\times {{10}^{-4}}\times 3.4\times {{10}^{-4}}}{1.7\times {{10}^{-5}}}\] \[=6.8\times {{10}^{-3}}\,M\]


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