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J & K CET Engineering J and K - CET Engineering Solved Paper-2013

  • question_answer
    The \[{{[\text{OH}]}^{-}}\]in a solution is \[1\,\text{mol}\,{{\text{L}}^{-1}}.\] The pH of the solution is

    A)  1      

    B)  0      

    C)  14     

    D)  \[{{10}^{-14}}\]

    Correct Answer: C

    Solution :

     Given, \[[O{{H}^{-}}]=1\,mol\,{{L}^{-1}}\] So, we know that \[pOH=-\log [O{{H}^{-}}]\] \[pOH=-\log \,1=0\] \[\because \] \[pH+pOH=14\] \[\therefore \] \[pH=14-0=14\]


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