2. Questions Bank
  3. JEE Main & Advanced
  4. Chemistry
  5. Equilibrium / साम्यावस्था

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

  • question_answer
    Assuming that the buffer in the blood is \[C{{O}_{2}}-HCO_{3}^{-}\]. Calculate the ratio of conjugate base to acid necessary to maintain blood at its proper pH of 7.4. \[{{K}_{1}}({{H}_{2}}CO{{  }_{3}})=4.5\times {{10}^{-7}}\]

    A) 11                    

    B) 8     

    C) 6                                 

    D) 14

    Correct Answer: A

    Solution :

    [a] \[C{{O}_{2}}\] with \[{{H}_{2}}O\] forms \[{{H}_{2}}C{{O}_{3}}\] \[C{{O}_{2}}+{{H}_{2}}O\rightleftharpoons {{H}^{+}}+HCO_{3}^{-}\] \[{{K}_{1}}=\frac{[{{H}^{+}}][HCO_{3}^{-}]}{[C{{O}_{2}}]}=4.5\times {{10}^{-7}}\] Again \[pH=-\log [{{H}^{+}}]=7.4\] \[\therefore \,[{{H}^{+}}]=4.0\times {{10}^{-8}}\] \[\therefore \,\frac{[HCO_{3}^{-}]}{[C{{O}_{2}}]}=\frac{4.5\times {{10}^{-7}}}{4\times {{10}^{-8}}}=11\]


You need to login to perform this action.
You will be redirected in 3 sec spinner