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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Acids and Bases

  • question_answer
    \[100\,\,ml\]of \[0.2\,\,M\,\,{{H}_{2}}S{{O}_{4}}\] is added to \[100\,\,ml\] of \[0.2\,\,M\,\,NaOH.\] The resulting solution will be   [BHU 1996]

    A)                 Acidic   

    B)                 Basic

    C)                 Neutral

    D)                 Slightly basic

    Correct Answer: A

    Solution :

               M.eq. of 0.2M \[{{H}_{2}}S{{O}_{4}}\]\[=\frac{2\times 0.2M}{1000}\times 100=0.04\] m/l                    M.eq. of .2M \[NaOH\]\[=\frac{0.2}{1000}\times 100=0.02\] m/l                    left \[[{{H}^{+}}]=.04-.02=.02\].                    Total volume\[=200=\frac{.02}{200}=.0001={{10}^{-4}}\]M                                 \[pH=4\].


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