JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer When \[C{{O}_{2}}\] dissolves in water, the following equilibrium is established \[C{{O}_{2}}+2{{H}_{2}}O{{H}_{3}}{{O}^{+}}+HC{{O}_{3}}^{-};\]for which the equilibrium constant is \[3.8\times {{10}^{-6}}\] and \[pH=6.0\] What would be the ratio of concentration of bicarbonate ion to carbon  dioxide?

    A)  \[3.8\times {{10}^{-12}}~\]

    B)  \[3.8\]                    

    C)   6                         

    D)   \[13.4\]

    Correct Answer: B

    Solution :

    \[K=\frac{[{{H}_{3}}{{O}^{+}}]\,[HC{{O}_{3}}^{-}]}{[C{{O}_{2}}]\,{{[{{H}_{2}}O]}^{2}}}\]As \[pH=6.0{{[{{H}_{3}}O]}^{+}}={{10}^{-6}}\] \[K=\frac{[{{H}_{3}}{{O}^{+}}]\,[HC{{O}_{3}}^{-}]}{[C{{O}_{2}}]\,{{[{{H}_{2}}O]}^{2}}}\] (\[{{H}_{2}}O\] is in excess, therefore its cone. remains constant) \[\frac{\left[ HC{{O}_{3}}^{-} \right]}{\left[ C{{O}_{2}} \right]}=\frac{K}{\left[ {{H}_{3}}{{O}^{+}} \right]}=\frac{3.8\times {{10}^{-6}}}{{{10}^{-6}}}=3.8\]


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