JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Self Evaluation Test - Equilibrium

  • question_answer
    At 298K a 0.1 M \[C{{H}_{2}}COOH\]solution is 1.34 % ionized. The ionization constant \[{{K}_{a}}\]for acetic acid will be

    A) \[1.82\times {{10}^{-5}}\]        

    B) \[18.2\times {{10}^{-5}}\]

    C) \[0.182\times {{10}^{-5}}\]     

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[{{K}_{a}}=c{{\alpha }^{2}}=0.1\times {{\left( \frac{1.34}{100} \right)}^{2}}=1.8\times {{10}^{-5}}\]


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