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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Self Evaluation Test - Equilibrium

  • question_answer
    If \[p{{K}_{b}}\] for fluoride ion at \[25{}^\circ C\] is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is

    A) \[3.52\times {{10}^{-3}}\]        

    B) \[6.75\times {{10}^{-4}}\]

    C) \[5.38\times {{10}^{-2}}\]        

    D) \[1.74\times {{10}^{-5}}\]

    Correct Answer: B

    Solution :

    [b] \[{{K}_{w}}={{K}_{a}}\times {{K}_{b}}\] \[{{K}_{b}}={{10}^{-10.83}}=1.48\times {{10}^{-11}}\] \[\therefore {{K}_{a}}=\frac{{{K}_{w}}}{{{K}_{b}}}=\frac{{{10}^{-14}}}{1.48\times {{10}^{-11}}}=6.75\times {{10}^{-4}}\]


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