A) \[1.0\times {{10}^{-5}}mol\,{{L}^{-1}}\]
B) \[1.0\times {{10}^{-6}}mol\,{{L}^{-1}}\]
C) \[2.0\times {{10}^{-6}}mol\text{ }{{L}^{-1}}\]
D) \[1.0\times {{10}^{-7}}mol\text{ }{{L}^{-1}}\]
Correct Answer: D
Solution :
[d] Given \[{{K}_{b}}=1.0\times {{10}^{-12}}\] \[\left[ BOH \right]=0.01\text{ }M\] \[{{[OH]}^{-}}=?\] \[\begin{align} & \begin{matrix} {} & {} & {} & BOH\rightleftharpoons {{B}^{+}}+O{{H}^{-}} \\ \end{matrix} \\ & \begin{matrix} t=0 & {} & c & \begin{matrix} {} & {} & 0 & \begin{matrix} {} & 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ & \begin{matrix} t={{t}_{eq}} & c\,(1-\alpha ) & \begin{matrix} c\alpha & c\alpha \\ \end{matrix} & {} \\ \end{matrix} \\ \end{align}\] \[{{K}_{b}}=\frac{{{c}^{2}}{{\alpha }^{2}}}{c(1-\alpha )}=\frac{c{{\alpha }^{2}}}{(1-\alpha )}\] \[\Rightarrow 1.0\times {{10}^{-12}}=\frac{0.01{{\alpha }^{2}}}{(1-\alpha )}\] On calculation, we get, \[\alpha =1.0\times {{10}^{-5}}\] Now, \[\left[ O{{H}^{-}} \right]=c\alpha =0.01\times {{10}^{-5}}\] \[=1\times {{10}^{-7}}mol\,{{L}^{-1}}\]
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