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JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    Dissociation constant of \[N{{H}_{4}}OH\] is \[1.8\times {{10}^{-5}}\]. The hydrolysis constant of \[N{{H}_{4}}Cl\] would be

    A)  \[1.8\times {{10}^{-19}}\]                           

    B)  \[5.55\times {{10}^{-10}}\]

    C)  \[5.55\times {{10}^{-5}}\]           

    D)  \[1.8\times {{10}^{-5}}\]

    Correct Answer: B

    Solution :

    Hydrolysis of \[N{{H}_{4}}Cl\] is written as \[NH_{4}^{+}+{{H}_{2}}ON{{H}_{4}}OH+{{H}^{+}}\] \[\Rightarrow {{K}_{h}}=\frac{{{K}_{w}}}{{{K}_{b}}}\] \[{{K}_{h}}=\frac{{{10}^{-14}}}{1.8\times {{10}^{-5}}}=0.5\times {{10}^{-9}}=5.5\times {{10}^{-10}}\]


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