JCECE Engineering JCECE Engineering 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 :

    \[HCO_{3}^{-}\] 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 hydrolysis 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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