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At $25^{\circ} \mathrm{C}$, 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
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$1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$
Base $B O H$ is dissociated as follows
$\mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-}$
So, the dissociation constant of $B O H$ base
$K_b=\frac{\left[B^{+}\right]\left[\mathrm{OH}^{-}\right]}{[B \mathrm{OH}]}...(i)$
At equilibrium $\left[B^{+}\right]=\left[\mathrm{OH}^{-}\right]$
$\therefore K_b=\frac{\left[\mathrm{OH}^{-}\right]^2}{[B \mathrm{OH}]}$
Given that $K_b=1.0 \times 10^{-12}$ and $\mathrm{BOH} = 0.01 \mathrm{~M}$
$\begin{aligned} \text{Therefore, } {\left[\mathrm{OH}^{-}\right]^2 } & =1 \times 10^{-14} \\ {\left[\mathrm{OH}^{-}\right] } & =1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$
$\mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-}$
So, the dissociation constant of $B O H$ base
$K_b=\frac{\left[B^{+}\right]\left[\mathrm{OH}^{-}\right]}{[B \mathrm{OH}]}...(i)$
At equilibrium $\left[B^{+}\right]=\left[\mathrm{OH}^{-}\right]$
$\therefore K_b=\frac{\left[\mathrm{OH}^{-}\right]^2}{[B \mathrm{OH}]}$
Given that $K_b=1.0 \times 10^{-12}$ and $\mathrm{BOH} = 0.01 \mathrm{~M}$
$\begin{aligned} \text{Therefore, } {\left[\mathrm{OH}^{-}\right]^2 } & =1 \times 10^{-14} \\ {\left[\mathrm{OH}^{-}\right] } & =1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$
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