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 :
\[BOH\rightleftharpoons {{B}^{+}}+O{{H}^{-}}\] Initial C 0 0 At eq. C ? Ca Ca Ca \[{{K}_{b}}=\frac{{{C}^{2}}{{\alpha }^{2}}}{C(1-\alpha )}=C{{\alpha }^{2}}\] assuming \[\alpha <<1\];\[1-\alpha \tilde{}1\] \[{{10}^{-12}}={{10}^{-2}}\times {{\alpha }^{2}}\]; \[{{\alpha }^{2}}={{10}^{-10}}\]; \[\alpha ={{10}^{-5}}\] \[[O{{H}^{-}}]=C\alpha =.01\times {{10}^{-5}}={{10}^{-7}}\]
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