JEE Main & Advanced Chemistry Equilibrium Question Bank Electrical conductors, Arrhenius theory and ostwald's dilution law

  • question_answer A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is      [BVP 2003]

    A)                 \[1\times {{10}^{-8}}\]  

    B)                 \[1\times {{10}^{-4}}\]

    C)                 \[1\times {{10}^{-6}}\]  

    D)                 \[{{10}^{-5}}\]

    Correct Answer: A

    Solution :

               \[K=\frac{{{\alpha }^{2}}C}{1-\alpha };\,\,\alpha =\frac{0.01}{100}\approx 1\,\,\,\therefore \,\,K={{\alpha }^{2}}C={{\left[ \frac{0.01}{100} \right]}^{2}}\times 1\] \[=1\times {{10}^{-8}}\].


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