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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    The pH of a \[0.02M\,N{{H}_{4}}Cl\]solution will be [given \[{{K}_{b}}(N{{H}_{4}}OH)={{10}^{-5}}\]and \[log2=0.301\]] [JEE Main 10-4-2019 Afternoon]

    A) 4.65

    B)              5.35

    C) 4.35                

    D) 2.65

    Correct Answer: B

    Solution :

    For the salt of strong acid and weak base \[{{H}^{+}}=\sqrt{\frac{{{K}_{\text{w}}}\times C}{{{K}_{b}}}}\] \[\left[ {{H}^{+}} \right]=\sqrt{\frac{{{10}^{-14}}\times 2\times {{10}^{-2}}}{{{10}^{-5}}}}\] \[-\log \left[ {{H}^{+}} \right]=6-\frac{1}{2}\log 20\] \[\therefore \]\[pH=5.35\]


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