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BCECE Engineering BCECE Engineering Solved Paper-2002

  • question_answer
    At \[25{{\,}^{o}}C\,{{K}_{a}}\]for\[~C{{H}_{3}}COOH\] is\[~1.8\times {{10}^{-5}}\]and \[{{K}_{b}}\]for \[\text{N}{{\text{H}}_{\text{4}}}\text{OH}\]is also \[\text{1}\text{.8}\times {{10}^{-5}}.\]The nature of aqueous solution of \[\text{C}{{\text{H}}_{\text{3}}}\text{COON}{{\text{H}}_{\text{4}}}\]is:

    A)  neutral                               

    B)  basic

    C) acidic                    

    D)        amphoteric

    Correct Answer: A

    Solution :

    Key Idea: Write hydrolysis reaction and then compare \[{{\text{K}}_{a}}\]and \[{{\text{K}}_{b}}\]to decide nature of resulting solution. \[\underset{\begin{smallmatrix}  \text{ammonimum} \\  \,\,\,\,\,\,\,\,\text{acetate} \end{smallmatrix}}{\mathop{C{{H}_{3}}COON{{H}_{4}}}}\,+{{H}_{2}}O\underset{\text{acetic}\,\text{acid}}{\mathop{C{{H}_{3}}COOH}}\,\]                                                \[+\,\underset{\begin{smallmatrix}  \text{ammonium} \\  \text{hydroxide} \end{smallmatrix}}{\mathop{N{{H}_{4}}OH}}\,\] Given \[{{K}_{a}}\]for \[C{{H}_{3}}COOH=1.8\times {{10}^{-5}}\] \[{{K}_{b}}\]for \[N{{H}_{4}}OH=1.8\times {{10}^{-5}}\] \[\because \]value of \[{{K}_{a}}\]and \[{{K}_{b}}\]is same. \[\therefore \] solution is neutral.


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