2. Solved Papers
  3. Punjab Medical

Punjab Medical Punjab - MET Solved Paper-2005

  • question_answer
    Degree of dissociation of \[N{{H}_{4}}OH\] in water is\[1.8\times {{10}^{-5}}\], then hydrolysis constant of \[N{{H}_{4}}Cl\] is:

    A) \[1.8\times {{10}^{-5}}\]                              

    B) \[1.8\times {{10}^{-10}}\]

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

    D) \[5.55\times {{10}^{-10}}\]

    Correct Answer: D

    Solution :

    We know\[{{K}_{H}}=\frac{{{K}_{w}}}{{{K}_{b}}}\] Here\[{{K}_{H}}=?\] \[{{K}_{w}}={{10}^{-14}},\,\,{{K}_{b}}=1.8\times {{10}^{-5}}\] \[{{K}_{H}}=\frac{{{10}^{-14}}}{1.8\times {{10}^{-5}}}=0.555\times {{10}^{-9}}\]        \[=5.55\times {{10}^{-10}}\]


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