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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Mock Test - Ionic Equilibrium

  • question_answer
    Calculate pH of 0.002 N \[N{{H}_{4}}\]OH having 2% dissociation.

    A) 7.6                   

    B) 8.6

    C) 9.6                  

    D) 10.6

    Correct Answer: C

    Solution :

    [c] \[N{{H}_{4}}OH\]is a weak base and partially dissociated  
      \[N{{H}_{4}}OH\rightleftharpoons N{{H}_{4}}^{+}+O{{H}^{-}}\]
    Concentration before dissociation  \[1\]   \[0\]     \[0\]
    Concentration after dissociation \[1-\alpha \]   \[\alpha \]    \[\alpha \]
    \[\therefore [O{{H}^{-}}]=C\alpha =2\times {{10}^{-3}}\times \frac{2}{100}=4\times {{10}^{-5}}M\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log 4\times {{10}^{-5}}=4.4\] \[pH=14-4.4=9.6\]


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