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

  • question_answer
    When 10 ml of 0.1 M acetic acid (\[p{{K}_{a}}\]. = 5.0) is titrated against 10ml of 0.l Mammonia solution (\[p{{K}_{b}}\]=5.0), the equivalence point occurs at pH

    A) 5.0                   

    B) 6.0

    C) 7.0                  

    D) 9.0

    Correct Answer: C

    Solution :

    [c] \[p{{K}_{a}}=-\log {{K}_{a}}p{{K}_{a,}}p{{K}_{b}}=-\log {{K}_{b}}\] \[pH=-\frac{1}{2}[log{{K}_{a}}+log{{K}_{w}}-log{{K}_{b}}]\] \[=-\frac{1}{2}[-5+log(1\times {{10}^{-14}})-(-5)]\] \[=-\frac{1}{2}[-5-14+5]=-\frac{1}{2}(-14)=7\]


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