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CMC Medical CMC-Medical VELLORE Solved Paper-2012

  • question_answer
    At \[25{}^\circ C\] the dissociation constant of a base BOH is\[1.0\times {{10}^{-12}}\]. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be

    A)  \[2.0\times {{10}^{-6}}mol\,{{L}^{-1}}\]

    B)  \[1.0\times {{10}^{-5}}mol\,{{L}^{-1}}\]

    C)  \[1.0\times {{10}^{-6}}mol\,{{L}^{-1}}\]

    D)  \[1.0\times {{10}^{-7}}mol\,{{L}^{-1}}\]

    Correct Answer: D

    Solution :

                    Base, \[BOH\] is dissociated as follows \[BOH{{B}^{+}}+O{{H}^{-}}\] So, the dissociation constant of \[BOH,\] \[{{K}_{b}}=\frac{[{{B}^{+}}]\,[O{{H}^{-}}]}{[BOH]}\] At equilibrium, \[[{{B}^{+}}]\,=[O{{H}^{-}}]\] \[\therefore \]  \[{{K}_{b}}=\frac{[O{{H}^{-}}]}{[BOH]}\] Given that   \[{{K}_{b}}=1.0\times {{10}^{-12}}\] and     \[[BOH]=0.01\,M\] Thus, \[1.0\times {{10}^{-12}}=\frac{{{[O{{H}^{-}}]}^{2}}}{0.01}\] \[{{[O{{H}^{-}}]}^{2}}=1\times {{10}^{-14}}\] \[[O{{H}^{-}}]=1.0\times {{10}^{-7}}mol\,{{L}^{-1}}\]


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