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 base \[{{K}_{b}}=\frac{[{{B}^{+}}]\,[O{{H}^{-}}]}{[BOH]}\] ?(i) At equilibrium \[[{{B}^{+}}]=[O{{H}^{-}}]\] \[\therefore \] \[{{K}_{b}}=\frac{{{[O{{H}^{-}}]}^{2}}}{[BOH]}\] 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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