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JEE Main & Advanced Sample Paper JEE Main Sample Paper-8

  • question_answer
    pH of the solution : \[\left( \begin{align}   & 10\,\underset{+}{\mathop{\,litre\,}}\,\,of\,\,0.03\,N\,X{{(OH)}_{2}}\,(strong\,\,diacidic\,\,base) \\  & 5\,\,\underset{+}{\mathop{litre}}\,\,\,0.08\,\,\,M\,\,HN{{O}_{3}} \\  & 485\,\,litre\,\,0.01\,\,M\,\,NaN{{O}_{3}} \\ \end{align} \right)\]is -

    A)  3.7                        

    B)  11.0

    C)  7.0                        

    D)  10.0

    Correct Answer: A

    Solution :

     \[{{[{{H}^{+}}]}_{remaining}}=\frac{5\times 0.08-10\times 0.03}{10+5+485}=2\times {{10}^{-4}}\] \[pH=3.7\]


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