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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Hydrogenion concentration pH scale and buffer solution

  • question_answer
    A buffer solution with \[pH\]9 is to be prepared by mixing \[N{{H}_{4}}Cl\] and \[N{{H}_{4}}OH\]. Calculate the number of moles of \[N{{H}_{4}}Cl\] that should be added to one litre of \[1.0\,M\ N{{H}_{4}}OH.\,\] \[[{{K}_{b}}=1.8\times {{10}^{-5}}]\]     [UPSEAT 2001]

    A)                 3.4         

    B)                 2.6

    C)                 1.5         

    D)                 1.8

    Correct Answer: D

    Solution :

               \[pH=-\text{log}\,\,{{K}_{b}}+\text{log}\frac{\text{ }\!\![\!\!\text{ Salt }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ Acid }\!\!]\!\!\text{ }}\]                    \[pH=-\text{log}\,[1.8\times {{10}^{-5}}]+\text{log}\frac{\text{ }\!\![\!\!\text{ Salt }\!\!]\!\!\text{ }}{\text{1}\text{.0}}\]                    \[9=4.7+\log \frac{\text{ }\!\![\!\!\text{ Salt }\!\!]\!\!\text{ }}{1.0}\] ; \[\log \frac{\text{ }\!\![\!\!\text{ Salt }\!\!]\!\!\text{ }}{1.0}=4.7-9=-4.3\]                                 \[\frac{\text{ }\!\![\!\!\text{ Salt }\!\!]\!\!\text{ }}{1.0}=\text{Antilog}\frac{1}{4.3};\,\,[S\text{alt }\!\!]\!\!\text{ }=1.8\]


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