JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer \[NaOH\] is a strong base. What will be pH of \[5.0\times {{10}^{-2}}\,M\,\,NaOH\] solution? \[(\log \,2\,=\,0.3)\]     JEE Main  Online Paper (Held On 22 April 2013)

    A)  14.00                                    

    B)  13.70

    C)  13.00                    

    D)  12.70

    Correct Answer: D

    Solution :

     Given \[[O{{H}^{-}}]=5\times {{10}^{-2}}\] \[\therefore \]\[pOH=-\log 5\times {{10}^{-2}}\] \[=-\log 5+2\log 10=1,30\] \[\because \]\[pH+pOH=14\] \[\because \]\[pH=14-pOH\]                 \[=14-1.30=12.70\]


You need to login to perform this action.
You will be redirected in 3 sec spinner