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NEET Chemistry Solutions / विलयन Question Bank Self Evaluation Test - Solutions

  • question_answer
    0.010M solution an acid HA freezes at \[-0.0205{}^\circ C\]. If\[~{{K}_{f}}\] for water is \[1.860K\text{ }kg\text{ }mo{{l}^{-1}}\], the ionization constant of the conjugate base of the acid will be (assume \[0.010M=0.010m\])  

    A) \[1.1\times {{10}^{-4}}\]         

    B) \[1.1\times {{10}^{-3}}\]

    C) \[9.0\times {{10}^{-11}}\]        

    D) \[9.0\times {{10}^{-12}}\]

    Correct Answer: C

    Solution :

    [c] \[\Delta {{T}_{f}}\left( normal \right)={{K}_{f}}m=1.86\times 0.01=0.0186;\] \[i=\frac{\Delta {{T}_{f(obs)}}}{\Delta {{T}_{f(nor)}}}=\frac{0.0205}{0.0186}=1.10=1+\alpha ;\] \[\alpha =0.1\] \[{{K}_{a}}=\frac{C{{a}^{2}}}{1-\alpha }=\frac{0.01\times {{0.1}^{2}}}{1-0.1}=\frac{1}{9}\times {{10}^{-3}};\] \[{{K}_{a}}=\frac{{{K}_{w}}}{{{K}_{a}}}=10\times {{10}^{-14}}\times 9\times {{10}^{3}}=9\times {{10}^{-11}}\]


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