Question

At $ 25{{\,}^{o}}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}} $

Answer 1

Answer: (B)

Base BOH is dissociated as follows $ BOH{{B}^{+}}+O{{H}^{-}} $
So, the dissociation constant of base BOH
$ {{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}} $