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JCECE Medical JCECE Medical Solved Paper-2013

  • question_answer
    Calculate pH of \[1\,M\,NaHC{{O}_{3}}.\]Given \[{{H}_{2}}C{{O}_{3}}+{{H}_{2}}OHCO_{3}^{-}+{{H}_{3}}{{O}^{+}};p{{K}_{1}}=6.38\]\[HCO_{3}^{-}+{{H}_{2}}OCO_{3}^{2-}+{{H}_{3}}{{O}^{+}};\,\,\,\,p{{K}_{2}}=10.26\]

    A)  8.73                

    B)  8.32,

    C)  6.73              

    D)  6.32

    Correct Answer: B

    Solution :

     \[\text{HCO}_{3}^{-}\]is proton acceptor as well as proton donor and therefore it is amphiprotic. For such case \[pH=\frac{p{{k}_{1}}+p{{k}_{2}}}{2}\] \[HCO_{3}^{-}\]undergoes hydrdysis and form alkaline solution. \[HCO_{3}^{-}+{{H}_{2}}O{{H}_{2}}C{{O}_{3}}+O{{H}^{-}}\] \[HCO_{3}^{-}\]can also ionise to form acidic solution. \[HCO_{3}^{-}+{{H}_{2}}O{{H}_{3}}{{O}^{+}}+CO_{3}^{2-}\] pH of the salt is, \[pH=\frac{p{{k}_{1}}+p{{k}_{2}}}{2}\] \[=\frac{6.38+10.26}{2}=8.32\]


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