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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2010

  • question_answer
    The pH of \[0.1\text{ }M\]aqueous solution of \[N{{H}_{4}}OH\]is \[({{K}_{b}}=1.0\times {{10}^{-5}})\]

    A)  \[3\]                                    

    B)  \[10.5\]

    C)  \[11\]                                  

    D)  \[7.5\]

    Correct Answer: C

    Solution :

    \[[O{{H}^{-}}]=\sqrt{{{K}_{b}}.C}=\sqrt{1.0\times {{10}^{-5}}\times 0.1}={{10}^{-3}}\] \[pOH=-\log [O{{H}^{-}}]=-\log \,{{10}^{-3}}=3\] \[pH=14-pOH=14-3=11\]


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