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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2014

  • question_answer
    For \[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O,\] \[{{E}^{o}}=1.33V\] At  \[[C{{r}_{2}}O_{7}^{2-}]=4.5\]mill mole,  \[[C{{r}^{3+}}]=15\]milli mole, E is\[1.067V\]. The pH of the solution is nearly equal to

    A)  \[2\]    

    B)  \[3\]     

    C)  \[5\]    

    D)  \[4\]

    Correct Answer: A

    Solution :

    : For the given reaction, \[C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\] \[E={{E}^{o}}-\frac{2.303RT}{nF}\log \frac{\text{ }\!\![\!\!\text{ Products }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ Reactants }\!\!]\!\!\text{ }}\] \[1.067=1.33-\frac{0.0591}{6}\log \frac{{{[C{{r}^{3+}}]}^{2}}{{[{{H}_{2}}O]}^{7}}}{[C{{r}_{2}}O_{7}^{2-}]{{[{{H}^{+}}]}^{14}}}\] \[1.067=1.33-9.85\times {{10}^{-3}}\log \frac{{{[15\times {{10}^{-3}}]}^{2}}{{[1]}^{7}}}{[4.5\times {{10}^{-3}}]{{[{{H}^{+}}]}^{14}}}\] \[\frac{-0.263}{-9.85\times {{10}^{-3}}}=2\log [15\times {{10}^{-3}}]-\log [4.5\times {{10}^{-3}}]\]  \[-14\log [{{H}^{+}}]\] \[26.7=2\times -1.82+2.34+14pH\] \[26.7=-3.64+2.34+14pH\] \[14pH=26.7+3.64-2.34\] \[pH=\frac{28}{14}=2\]


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