Equilibrium Question 827
Question: 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
Options:
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{L^{-1}} $
D) $ 1.0\times {10^{-7}}mol{L^{-1}} $
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Answer:
Correct Answer: D
Solution:
Given $ K_{b}=1.0\times {10^{-12}} $
$ [ BOH ]=0.01M $
$ {{[OH]}^{-}}=- $
$ \begin{aligned} & \begin{matrix} {} & {} & {} & BOH\rightarrow {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{aligned} $
$ 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, $ [ O{H^{-}} ]=c\alpha =0.01\times {10^{-5}} $
$ =1\times {10^{-7}}mol,{L^{-1}} $
